LaTeX vs MathPix image — Introduction to Linear and Matrix Algebra (Nathaniel Johnston) (Z-Library)

\mathbf{v}+\mathbf{w} \stackrel{\text { def }}{=}\left(v_{1}+w_{1}, v_{2}+w_{2}, \ldots, v_{n}+w_{n}\right) .

\begin{aligned} \mathbf{v}+\mathbf{w} & =\left(v_{1}+w_{1}, v_{2}+w_{2}, \ldots, v_{n}+w_{n}\right) \\ & =\left(w_{1}+v_{1}, w_{2}+v_{2}, \ldots, w_{n}+v_{n}\right)=\mathbf{w}+\mathbf{v} . \end{aligned}

c \mathbf{v} \stackrel{\text { def }}{=}\left(c v_{1}, c v_{2}, \ldots, c v_{n}\right) .

\begin{aligned} c(\mathbf{v}+\mathbf{w}) & =c\left(v_{1}+w_{1}, v_{2}+w_{2}, \ldots, v_{n}+w_{n}\right) & & \text { (vector addition) } \\ & =\left(c\left(v_{1}+w_{1}\right), c\left(v_{2}+w_{2}\right), \ldots, c\left(v_{n}+w_{n}\right)\right) & & \text { (scalar mult.) } \\ & =\left(c v_{1}+c w_{1}, c v_{2}+c w_{2}, \ldots, c v_{n}+c w_{n}\right) & & \text { (property of } \mathbb{R} \text { ) } \\ & =\left(c v_{1}, c v_{2}, \ldots, c v_{n}\right)+\left(c w_{1}, c w_{2}, \ldots, c w_{n}\right) & & \text { (vector addition) } \\ & =c\left(v_{1}, v_{2}, \ldots, v_{n}\right)+c\left(w_{1}, w_{2}, \ldots, w_{n}\right) & & \text { (scalar mult.) } \\ & =c \mathbf{v}+c \mathbf{w} . & & \end{aligned}

\begin{array}{rlrl} & & \mathbf{x}-(3,2,1) & =(1,2,3)-3 \mathbf{x} \\ \Longrightarrow & \mathbf{x} & =(4,4,4)-3 \mathbf{x} & \text { (add }(3,2,1) \text { to both sides) } \\ \Longrightarrow & 4 \mathbf{x} & =(4,4,4) & \\ \Longrightarrow & \mathbf{x} & =(1,1,1) . & \\ \Longrightarrow & \text { (divide both sides by } 4) \end{array}

\begin{array}{rlrl} & & \mathbf{x}+2(\mathbf{v}+\mathbf{w}) & =-\mathbf{v}-3(\mathbf{x}-\mathbf{w}) \\ \Longrightarrow & \mathbf{x}+2 \mathbf{v}+2 \mathbf{w} & =-\mathbf{v}-3 \mathbf{x}+3 \mathbf{w} & \\ \Longrightarrow & 4 \mathbf{x} & =-3 \mathbf{v}+\mathbf{w} & \\ \Longrightarrow & \mathbf{x} & =\frac{1}{4}(\mathbf{w}-3 \mathbf{v}) . & \\ \Longrightarrow & \text { (divide both sides by } 4) \end{array}

c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k},

\mathbf{e}_{j} \stackrel{\text { def }}{=}(0,0, \ldots, 0,1,0, \ldots, 0) .

\mathbf{v}=v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n},

\mathbf{v w} \stackrel{\text { def }}{=}\left(v_{1} w_{1}, v_{2} w_{2}, \ldots, v_{n} w_{n}\right) .

\mathbf{v} \cdot \mathbf{w} \stackrel{\text { def }}{=} v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n} .

\begin{aligned} \left(v_{1}, v_{2}, \ldots, v_{n}\right) \cdot \mathbf{e}_{j} & =0 v_{1}+\cdots+0 v_{j-1}+1 v_{j}+0 v_{j+1}+\cdots+0 v_{n} \\ & =v_{j} \end{aligned}

\begin{aligned} \mathbf{v} \cdot \mathbf{w} & =v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n} \\ & =w_{1} v_{1}+w_{2} v_{2}+\cdots+w_{n} v_{n}=\mathbf{w} \cdot \mathbf{v} . \end{aligned}

\left.\begin{array}{rl} (\mathbf{v}+\mathbf{w}) \cdot(\mathbf{x}+\mathbf{y}) & =(\mathbf{v}+\mathbf{w}) \cdot \mathbf{x}+(\mathbf{v}+\mathbf{w}) \cdot \mathbf{y} \\ & =\mathbf{x} \cdot(\mathbf{v}+\mathbf{w})+\mathbf{y} \cdot(\mathbf{v}+\mathbf{w}) \\ & =\mathbf{x} \cdot \mathbf{v}+\mathbf{x} \cdot \mathbf{w}+\mathbf{y} \cdot \mathbf{v}+\mathbf{y} \cdot \mathbf{w} \\ & =\mathbf{v} \cdot \mathbf{x}+\mathbf{w} \cdot \mathbf{x}+\mathbf{v} \cdot \mathbf{y}+\mathbf{w} \cdot \mathbf{y} \end{array}(\text { property (broperty (a) })\right)

\|\mathbf{v}\|=\sqrt{\left\|\left(v_{1}, 0\right)\right\|^{2}+\left\|\left(0, v_{2}\right)\right\|^{2}}=\sqrt{\left|v_{1}\right|^{2}+\left|v_{2}\right|^{2}}=\sqrt{v_{1}^{2}+v_{2}^{2}}=\sqrt{\mathbf{v} \cdot \mathbf{v}} .

\begin{aligned} \|\mathbf{v}\| & =\sqrt{\left\|\left(v_{1}, v_{2}, 0\right)\right\|^{2}+\left\|\left(0,0, v_{3}\right)\right\|^{2}} \\ & =\sqrt{\left(\sqrt{v_{1}^{2}+v_{2}^{2}}\right)^{2}+\left|v_{3}\right|^{2}}=\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}}=\sqrt{\mathbf{v} \cdot \mathbf{v}} . \end{aligned}

\|\mathbf{v}\| \stackrel{\text { def }}{=} \sqrt{\mathbf{v} \cdot \mathbf{v}}=\sqrt{v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}} .

\begin{aligned} \|c \mathbf{v}\| & =\sqrt{\left(c v_{1}\right)^{2}+\left(c v_{2}\right)^{2}+\cdots+\left(c v_{n}\right)^{2}} \\ & =\sqrt{c^{2}\left(v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}\right)} \\ & =\sqrt{c^{2}} \sqrt{v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}}=|c|\|\mathbf{v}\| . \end{aligned}

\left\|\frac{\mathbf{v}}{\|\mathbf{v}\|}\right\|=\frac{1}{\|\mathbf{v}\|}\|\mathbf{v}\|=1 .

\begin{aligned} 0 \leq\|c \mathbf{v}+d \mathbf{w}\|^{2} & =(c \mathbf{v}+d \mathbf{w}) \cdot(c \mathbf{v}+d \mathbf{w}) \\ & =c^{2}(\mathbf{v} \cdot \mathbf{v})+2 c d(\mathbf{v} \cdot \mathbf{w})+d^{2}(\mathbf{w} \cdot \mathbf{w}) \\ & =c^{2}\|\mathbf{v}\|^{2}+2 c d(\mathbf{v} \cdot \mathbf{w})+d^{2}\|\mathbf{w}\|^{2} \end{aligned}

\begin{aligned} 0 & \leq\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}-2\|\mathbf{w}\|(\mathbf{v} \cdot \mathbf{w})^{2} /\|\mathbf{w}\|+(\mathbf{v} \cdot \mathbf{w})^{2}\|\mathbf{w}\|^{2} /\|\mathbf{w}\|^{2} \\ & =\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}-(\mathbf{v} \cdot \mathbf{w})^{2} . \end{aligned}

\begin{aligned} c^{2}= & a^{2}+b^{2} \\ & -2 a b \cos (\theta) . \end{aligned}

\begin{aligned} \|\mathbf{v}+\mathbf{w}\|^{2} & =(\mathbf{v}+\mathbf{w}) \cdot(\mathbf{v}+\mathbf{w}) & & \text { (definition of length) } \\ & =(\mathbf{v} \cdot \mathbf{v})+2(\mathbf{v} \cdot \mathbf{w})+(\mathbf{w} \cdot \mathbf{w}) & & \text { (dot product properties (FOIL)) } \\ & =\|\mathbf{v}\|^{2}+2(\mathbf{v} \cdot \mathbf{w})+\|\mathbf{w}\|^{2} & & \text { (definition of length) } \\ & \leq\|\mathbf{v}\|^{2}+2\|\mathbf{v}\|\|\mathbf{w}\|+\|\mathbf{w}\|^{2} & & \text { (Cauchy-Schwarz inequality) } \\ & =(\|\mathbf{v}\|+\|\mathbf{w}\|)^{2} & & \text { (factor cleverly) } \end{aligned}

\|\mathbf{v}-\mathbf{w}\|^{2}=\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}-2\|\mathbf{v}\|\|\mathbf{w}\| \cos (\theta) .

\begin{aligned} \|\mathbf{v}-\mathbf{w}\|^{2} & =(\mathbf{v}-\mathbf{w}) \cdot(\mathbf{v}-\mathbf{w}) \\ & =\mathbf{v} \cdot \mathbf{v}-\mathbf{v} \cdot \mathbf{w}-\mathbf{w} \cdot \mathbf{v}+\mathbf{w} \cdot \mathbf{w}=\|\mathbf{v}\|^{2}-2(\mathbf{v} \cdot \mathbf{w})+\|\mathbf{w}\|^{2} . \end{aligned}

\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}-2\|\mathbf{v}\|\|\mathbf{w}\| \cos (\theta)=\|\mathbf{v}\|^{2}-2(\mathbf{v} \cdot \mathbf{w})+\|\mathbf{w}\|^{2} .

\cos (\theta)=\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\|\|\mathbf{w}\|}, \quad \text { so } \quad \theta=\arccos \left(\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\|\|\mathbf{w}\|}\right) .

\theta=\arccos \left(\frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\|\|\mathbf{w}\|}\right) .

\theta=\arccos \left(\frac{11}{5 \sqrt{5}}\right) \approx 0.1799 \text { radians (or } \approx 10.30 \text { degrees). }

\theta=\arccos (0)=\pi / 2 \text { (i.e., } 90 \text { degrees). }

\theta=\arccos \left(\frac{1}{\sqrt{2} \cdot \sqrt{2}}\right)=\arccos \left(\frac{1}{2}\right)=\pi / 3 \text { (i.e., } 60 \text { degrees). }

\left(\frac{v_{1}+v_{2}+\cdots+v_{n}}{n}\right)^{2} \leq \frac{1}{n}\left(v_{1}^{2}+v_{2}^{2}+\cdots+v_{n}^{2}\right) .

\mathbf{v} \cdot \mathbf{w} \stackrel{\text { def }}{=} \overline{v_{1}} w_{1}+\overline{v_{2}} w_{2}+\cdots+\overline{v_{n}} w_{n},

A=\left[\begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ccc} 3 & 0 & 2 \\ 0 & -1 & 1 \end{array}\right] .

\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 \end{array}\right] \text { and }\left[\begin{array}{cccc} 2 & -3 & & 4 \\ & 1 & 2 & 0 \end{array}\right] .

[A+B]_{i, j}=a_{i, j}+b_{i, j} \quad \text { and } \quad[c A]_{i, j}=c a_{i, j},

\begin{aligned} A+B & =\left[\begin{array}{cccc} a_{1,1}+b_{1,1} & a_{1,2}+b_{1,2} & \cdots & a_{1, n}+b_{1, n} \\ a_{2,1}+b_{2,1} & a_{2,2}+b_{2,2} & \cdots & a_{2, n}+b_{2, n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m, 1}+b_{m, 1} & a_{m, 2}+b_{m, 2} & \cdots & a_{m, n}+b_{m, n} \end{array}\right] \\ & =\left[\begin{array}{cccc} b_{1,1}+a_{1,1} & b_{1,2}+a_{1,2} & \cdots & b_{1, n}+a_{1, n} \\ b_{2,1}+a_{2,1} & b_{2,2}+a_{2,2} & \cdots & b_{2, n}+a_{2, n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{m, 1}+a_{m, 1} & b_{m, 2}+a_{m, 2} & \cdots & b_{m, n}+a_{m, n} \end{array}\right]=B+A, \end{aligned}

[A B]_{i, j} \stackrel{\text { def }}{=} a_{i, 1} b_{1, j}+a_{i, 2} b_{2, j}+\cdots+a_{i, n} b_{n, j} .

A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{ccc} 5 & 6 & 7 \\ 8 & 9 & 10 \end{array}\right], \quad \text { and } \quad C=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ 2 & -1 \end{array}\right] .

\begin{aligned} A B & =\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{lll} 5 & 6 & 7 \\ 8 & 9 & 10 \end{array}\right] \\ & =\left[\begin{array}{ccc} (1,2) \cdot(5,8) & (1,2) \cdot(6,9) & (1,2) \cdot(7,10) \\ (3,4) \cdot(5,8) & (3,4) \cdot(6,9) & (3,4) \cdot(7,10) \end{array}\right] \\ & =\left[\begin{array}{lll} 1 \times 5+2 \times 8 & 1 \times 6+2 \times 9 & 1 \times 7+2 \times 10 \\ 3 \times 5+4 \times 8 & 3 \times 6+4 \times 9 & 3 \times 7+4 \times 10 \end{array}\right] \\ & =\left[\begin{array}{lll} 21 & 24 & 27 \\ 47 & 54 & 61 \end{array}\right] . \end{aligned}

\begin{aligned} B C & =\left[\begin{array}{ccc} 5 & 6 & 7 \\ 8 & 9 & 10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ 2 & -1 \end{array}\right] \\ & =\left[\begin{array}{cc} (5,6,7) \cdot(1,0,2) & (5,6,7) \cdot(0,-1,-1) \\ (8,9,10) \cdot(1,0,2) & (8,9,10) \cdot(0,-1,-1) \end{array}\right] \\ & =\left[\begin{array}{cc} 5 \times 1+6 \times 0+7 \times 2 & 5 \times 0+6 \times-1+7 \times-1 \\ 8 \times 1+9 \times 0+10 \times 2 & 8 \times 0+9 \times-1+10 \times-1 \end{array}\right] \\ & =\left[\begin{array}{cc} 19 & -13 \\ 28 & -19 \end{array}\right] . \end{aligned}

a_{i, 1}\left(b_{1, j}+c_{1, j}\right)+a_{i, 2}\left(b_{2, j}+c_{2, j}\right)+\cdots+a_{i, n}\left(b_{n, j}+c_{n, j}\right),

\left(a_{i, 1} b_{1, j}+a_{i, 2} b_{2, j}+\cdots+a_{i, n} b_{n, j}\right)+\left(a_{i, 1} c_{1, j}+a_{i, 2} c_{2, j}+\cdots+a_{i, n} c_{n, j}\right) .

A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right]

A B=\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right] \quad \text { and } \quad B A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] .

I_{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad \text { and } \quad I_{3}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] .

A^{k} \stackrel{\text { def }}{=} \underbrace{A A \cdots A}_{k \text { copies }} .

A^{2}=\left[\begin{array}{cc} 2 & 1 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 1 \\ -1 & 3 \end{array}\right]=\left[\begin{array}{cc} 3 & 5 \\ -5 & 8 \end{array}\right] .

A^{4}=\left(A^{2}\right)^{2}=\left[\begin{array}{cc} 3 & 5 \\ -5 & 8 \end{array}\right]\left[\begin{array}{cc} 3 & 5 \\ -5 & 8 \end{array}\right]=\left[\begin{array}{cc} -16 & 55 \\ -55 & 39 \end{array}\right] .

\begin{aligned} (A+B)^{2} & =(A+B)(A+B) \\ & =A(A+B)+B(A+B) \\ & =A A+A B+B A+B B=A^{2}+A B+B A+B^{2}, \end{aligned}

\begin{aligned} (A+I)^{2}(A-I) & =\left(A^{2}+A I+I A+I\right)(A-I) \\ & =\left(A^{2}+2 A+I\right)(A-I) \\ & =\left(A^{3}+2 A^{2}+A\right)-\left(A^{2}+2 A+I\right) \\ & =A^{3}+A^{2}-A-I . \end{aligned}

\text { if } \quad A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad \text { and } \quad \mathbf{v}=\left[\begin{array}{c} 2 \\ -1 \end{array}\right] \quad \text { then } \quad A \mathbf{v}=\left[\begin{array}{l} 2-2 \\ 6-4 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \end{array}\right],

\mathbf{w}=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \in \mathcal{M}_{3,1}

A=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] \quad \Longrightarrow \quad A^{T}=\left[\begin{array}{cc} 7 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}\right] .

A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right] .

\left[(A B)^{T}\right]_{i, j}=[A B]_{j, i}=a_{j, 1} b_{1, i}+a_{j, 2} b_{2, i}+\cdots+a_{j, n} b_{n, i},

\begin{aligned} {\left[B^{T} A^{T}\right]_{i, j} } & =\left[B^{T}\right]_{i, 1}\left[A^{T}\right]_{1, j}+\left[B^{T}\right]_{i, 2}\left[A^{T}\right]_{2, j}+\cdots+\left[B^{T}\right]_{i, n}\left[A^{T}\right]_{n, j} \\ & =b_{1, i} a_{j, 1}+b_{2, i} a_{j, 2}+\cdots+b_{n, i} a_{j, n} . \end{aligned}

\mathbf{v}^{T} \mathbf{w}=\left[\begin{array}{llll} v_{1} & v_{2} & \cdots & v_{n} \end{array}\right]\left[\begin{array}{c} w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{array}\right]=v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n}=\mathbf{v} \cdot \mathbf{w} .

A=\left[\begin{array}{cccccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 & 1 & -1 \\ 0 & 0 & 0 & 0 & -2 & 3 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cccc} 1 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & -1 & 1 \end{array}\right]

A=\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \hline 0 & 0 & 0 & 2 & 1 & -1 \\ 0 & 0 & 0 & 0 & -2 & 3 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cc|cc} 1 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 2 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & -1 & 1 \end{array}\right] .

C=\left[\begin{array}{ccc} 2 & 1 & -1 \\ 0 & -2 & 3 \end{array}\right] \quad \text { and } \quad D=\left[\begin{array}{cc} 1 & 2 \\ 2 & 1 \\ -1 & 1 \end{array}\right],

A=\left[\begin{array}{cc} I_{3} & I_{3} \\ O & C \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cc} D & O \\ O & D \end{array}\right] .

\begin{aligned} A B & =\left[\begin{array}{ll} I_{3} & I_{3} \\ O & C \end{array}\right]\left[\begin{array}{ll} D & O \\ O & D \end{array}\right] \\ & \left.=\left[\begin{array}{ll} I_{3} & D+I_{3} \end{array}\right] \quad \text { O } \quad \begin{array}{ll} I_{3} & O+I_{3} \\ O & D+C \\ O \end{array} \quad \begin{array}{ll} O & O+C \\ O \end{array}\right]=\left[\begin{array}{cc} D & D \\ O & C D \end{array}\right] . \end{aligned}

A B=\left[\begin{array}{cc} D & D \\ O & C D \end{array}\right]=\left[\begin{array}{cc|cc} 1 & 2 & 1 & 2 \\ 2 & 1 & 2 & 1 \\ -1 & 1 & -1 & 1 \\ \hline 0 & 0 & 5 & 4 \\ 0 & 0 & -7 & 1 \end{array}\right],

A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right] .

\left[\begin{array}{ll} A & B \\ B & A \end{array}\right]=\left[\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 1 \\ 3 & 4 & 0 & 1 & 0 \\ \hline-1 & 1 & 1 & 1 & 2 & \\ 0 & 1 & 0 & 3 & 4 & \end{array}\right],

\left[\begin{array}{ll} A & B \\ O & I_{3} \end{array}\right]\left[\begin{array}{ll} A & A \\ O & A \end{array}\right]=\left[\begin{array}{cc} A^{2} & A^{2}+B A \\ O & A \end{array}\right],

\left[\begin{array}{ll} A & B \\ O & I_{3} \end{array}\right]\left[\begin{array}{ll} B & O \\ I_{3} & I_{3} \end{array}\right]=\left[\begin{array}{cc} A B+B & B \\ I_{3} & I_{3} \end{array}\right]=\left[\begin{array}{ccc|ccc} -2 & 4 & 2 & -1 & 1 & 1 \\ -3 & 8 & 3 & 0 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right] .

A \mathbf{v}=v_{1} \mathbf{a}_{1}+v_{2} \mathbf{a}_{2}+\cdots+v_{n} \mathbf{a}_{n} .

A \mathbf{v}=\left[\begin{array}{l|l|l|l} \mathbf{a}_{1} & \mathbf{a}_{2} & \cdots & \mathbf{a}_{n} \end{array}\right]\left[\begin{array}{c} v_{1} \\ v_{2} \\ \vdots \\ v_{n} \end{array}\right]=v_{1} \mathbf{a}_{1}+v_{2} \mathbf{a}_{2}+\cdots+v_{n} \mathbf{a}_{n},

A B=\left[A \mathbf{b}_{1}\left|A \mathbf{b}_{2}\right| \cdots \mid A \mathbf{b}_{p}\right] .

A B=A\left[\mathbf{b}_{1}\left|\mathbf{b}_{2}\right| \cdots \mid \mathbf{b}_{p}\right]=\left[A \mathbf{b}_{1}\left|A \mathbf{b}_{2}\right| \cdots \mid A \mathbf{b}_{p}\right],

\begin{aligned} A B & =\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{ccc} -1 & 1 & 1 \\ 0 & 1 & 0 \end{array}\right] \\ & =\left[\begin{array}{ccc} (1,2) \cdot(-1,0) & (1,2) \cdot(1,1) & (1,2) \cdot(1,0) \\ (3,4) \cdot(-1,0) & (3,4) \cdot(1,1) & (3,4) \cdot(1,0) \end{array}\right] \\ & =\left[\begin{array}{ccc} -1 & 3 & 1 \\ -3 & 7 & 3 \end{array}\right] . \end{aligned}

\mathbf{b}_{1}=\left[\begin{array}{c} -1 \\ 0 \end{array}\right], \quad \mathbf{b}_{2}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right], \quad \text { and } \quad \mathbf{b}_{3}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] .

A \mathbf{b}_{1}=\left[\begin{array}{c} -1 \\ -3 \end{array}\right], \quad A \mathbf{b}_{2}=\left[\begin{array}{l} 3 \\ 7 \end{array}\right], \quad \text { and } \quad A \mathbf{b}_{3}=\left[\begin{array}{l} 1 \\ 3 \end{array}\right] .

A B=\left[\begin{array}{lll} -1 & 3 & 1 \\ -3 & 7 & 3 \end{array}\right] .

A=\left[\begin{array}{cc} 2 & -1 \\ 0 & 3 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cc} -1 & 1 \\ -2 & 0 \end{array}\right] .

C=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 1 & 0 & 1 \end{array}\right] \quad \text { and } \quad D=\left[\begin{array}{ccc} 0 & -2 & 1 \\ 2 & 2 & 0 \end{array}\right] .

A=\left[\begin{array}{cc} 0.8 & 0.6 \\ 0.2 & h \end{array}\right] .

J_{2}=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right] \quad \text { and } \quad J_{3}=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] .

A=\left[\begin{array}{lllllll} 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 \end{array}\right] .

A B=\left[\begin{array}{c} \frac{\mathbf{a}_{1}^{T} B}{\mathbf{a}_{2}^{T} B} \\ \hline \vdots \\ \hline \mathbf{a}_{m}^{T} B \end{array}\right] .

T\left(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}\right)=c_{1} T\left(\mathbf{v}_{1}\right)+c_{2} T\left(\mathbf{v}_{2}\right)+\cdots+c_{k} T\left(\mathbf{v}_{k}\right)

\begin{aligned} T(\mathbf{v}+\mathbf{w}) & =T\left(v_{1}+w_{1}, v_{2}+w_{2}\right) \\ & =\left(\left(v_{1}+w_{1}\right)-\left(v_{2}+w_{2}\right),\left(v_{1}+w_{1}\right)+\left(v_{2}+w_{2}\right)\right) \\ & =\left(v_{1}-v_{2}, v_{1}+v_{2}\right)+\left(w_{1}-w_{2}, w_{1}+w_{2}\right)=T(\mathbf{v})+T(\mathbf{w}), \end{aligned}

\begin{aligned} T(c \mathbf{v})=T\left(c v_{1}, c v_{2}\right) & =\left(c v_{1}-c v_{2}, c v_{1}+c v_{2}\right) \\ & =c\left(v_{1}-v_{2}, v_{1}+v_{2}\right)=c T(\mathbf{v}) . \end{aligned}

T(\mathbf{v})=T\left(v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n}\right)=v_{1} T\left(\mathbf{e}_{1}\right)+v_{2} T\left(\mathbf{e}_{2}\right)+\cdots+v_{n} T\left(\mathbf{e}_{n}\right),

\begin{aligned} T\left(v_{1}, v_{2}\right) & =T\left(v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}\right) \\ & =v_{1} T\left(\mathbf{e}_{1}\right)+v_{2} T\left(\mathbf{e}_{2}\right) \\ & =v_{1}(1,1)+v_{2}(-1,1)=\left(v_{1}-v_{2}, v_{1}+v_{2}\right) . \end{aligned}

\text { if } \quad A=\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right] \quad \text { then } \quad A \mathbf{v}=\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right]=\left[\begin{array}{l} v_{1}-v_{2} \\ v_{1}+v_{2} \end{array}\right],

T(\mathbf{v})=[T] \mathbf{v} \quad \text { for all } \quad \mathbf{v} \in \mathbb{R}^{n} .

[T] \stackrel{\text { def }}{=}\left[T\left(\mathbf{e}_{1}\right)\left|T\left(\mathbf{e}_{2}\right)\right| \cdots \mid T\left(\mathbf{e}_{n}\right)\right] .

[T]=\left[T\left(\mathbf{e}_{1}\right)\left|T\left(\mathbf{e}_{2}\right)\right| \cdots \mid T\left(\mathbf{e}_{n}\right)\right]

\begin{aligned} {[T] \mathbf{v} } & =\left[T\left(\mathbf{e}_{1}\right)\left|T\left(\mathbf{e}_{2}\right)\right| \cdots \mid T\left(\mathbf{e}_{n}\right)\right]\left[\begin{array}{c} v_{1} \\ v_{2} \\ \vdots \\ v_{n} \end{array}\right] & & \\ & =v_{1} T\left(\mathbf{e}_{1}\right)+v_{2} T\left(\mathbf{e}_{2}\right)+\cdots+v_{n} T\left(\mathbf{e}_{n}\right) & & \text { (block matrix multiplication) } \\ & =T\left(v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n}\right) & & (\text { since } T \text { is linear }) \\ & =T(\mathbf{v}) . & & \left(\mathbf{v}=v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n}\right) \end{aligned}

[T]=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] .

[T]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ 2 & 4 & -2 \end{array}\right] .

(S+T)(\mathbf{v})=S(\mathbf{v})+T(\mathbf{v}) \quad \text { and } \quad(c T)(\mathbf{v})=c T(\mathbf{v}) \quad \text { for all } \quad \mathbf{v} \in \mathbb{R}^{n},

[T]=\left[c_{1} \mathbf{e}_{1}\left|c_{2} \mathbf{e}_{2}\right| \cdots \mid c_{n} \mathbf{e}_{n}\right]=\left[\begin{array}{cccc} c_{1} & 0 & \cdots & 0 \\ 0 & c_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & c_{n} \end{array}\right] .

\left[\begin{array}{cccc} c_{1} & 0 & \cdots & 0 \\ 0 & c_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & c_{n} \end{array}\right]\left[\begin{array}{cccc} d_{1} & 0 & \cdots & 0 \\ 0 & d_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_{n} \end{array}\right]=\left[\begin{array}{cccc} c_{1} d_{1} & 0 & \cdots & 0 \\ 0 & c_{2} d_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & c_{n} d_{n} \end{array}\right] .

\begin{array}{lll} {\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]} & =\left[\begin{array}{lll} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{array}\right] \text { and } \\ {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right]} & =\left[\begin{array}{lll} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{array}\right] \end{array}

P_{\mathbf{u}}(\mathbf{v})=\mathbf{u}(\|\mathbf{v}\|(\mathbf{u} \cdot \mathbf{v}) /\|\mathbf{v}\|)=\mathbf{u}(\mathbf{u} \cdot \mathbf{v}) .

\left[P_{\mathbf{u}}\right]=\mathbf{u} \mathbf{u}^{T} .

\left[P_{\mathbf{u}}\right]=\mathbf{u} \mathbf{u}^{T}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] .

\left[P_{\mathbf{u}}\right] \mathbf{v}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right]=\left[\begin{array}{c} v_{1} \\ 0 \end{array}\right] .

\left[P_{\mathbf{u}}\right]=\mathbf{u u}^{T}=\frac{1}{14}\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right]=\frac{1}{14}\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right] .

\left[P_{\mathbf{u}}\right] \mathbf{w}=\frac{1}{14}\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]=\mathbf{w} .

\begin{aligned} & 2\left(\mathbf{u} \mathbf{u}^{T}\right) \mathbf{v}-\mathbf{v} \\ & \quad=\left(2 \mathbf{u} \mathbf{u}^{T}-1\right) \mathbf{v}, \end{aligned}

\begin{aligned} F_{\mathbf{u}}(\mathbf{v}) & =\mathbf{v}+2\left(P_{\mathbf{u}}(\mathbf{v})-\mathbf{v}\right) & & (\text { by Figure 1.17 }) \\ & =2 P_{\mathbf{u}}(\mathbf{v})-\mathbf{v} & & (\text { expand parentheses }) \\ & =2\left(\mathbf{u u}^{T}\right) \mathbf{v}-\mathbf{v} & & \left(\text { since } P_{\mathbf{u}}(\mathbf{v})=\left(\mathbf{u u}^{T}\right) \mathbf{v}\right) \\ & =\left(2 \mathbf{u u}^{T}-I\right) \mathbf{v} . & & (\text { factor carefully }) \end{aligned}

\left[F_{\mathbf{u}}\right]=2 \mathbf{u} \mathbf{u}^{T}-I .

\left[F_{\mathbf{u}}\right]=2 \mathbf{u} \mathbf{u}^{T}-I=2\left[\begin{array}{l} 0 \\ 1 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right] .

\left[F_{\mathbf{u}}\right] \mathbf{v}=\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right]=\left[\begin{array}{c} -v_{1} \\ v_{2} \end{array}\right] .

\begin{aligned} {\left[F_{\mathbf{u}}\right]=2 \mathbf{u u}^{T}-I } & =2\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]\left[\begin{array}{lll} 1 & 1 & 1 \end{array}\right] / 3-\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & =\frac{1}{3}\left[\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\ 2 & 2 & -1 \end{array}\right] . \end{aligned}

\left[F_{\mathbf{u}}\right] \mathbf{w}=\frac{1}{3}\left[\begin{array}{ccc} -1 & 2 & 2 \\ 2 & -1 & 2 \\ 2 & 2 & -1 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]=\mathbf{w} .

\left[F_{\mathbf{u}}\right]=2 \mathbf{u} \mathbf{u}^{T}-I=\frac{1}{2}\left[\begin{array}{c} 1 \\ \sqrt{3} \end{array}\right]\left[\begin{array}{cc} 1 & \sqrt{3} \end{array}\right]-\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc} -1 & \sqrt{3} \\ \sqrt{3} & 1 \end{array}\right] .

\left[F_{\mathbf{u}}\right] \mathbf{v}=\frac{1}{2}\left[\begin{array}{cc} -1 & \sqrt{3} \\ \sqrt{3} & 1 \end{array}\right]\left[\begin{array}{c} -1 \\ 3 \end{array}\right]=\frac{1}{2}\left[\begin{array}{c} 3 \sqrt{3}+1 \\ 3-\sqrt{3} \end{array}\right] \approx\left[\begin{array}{c} 3.0981 \\ 0.6340 \end{array}\right] .

\begin{figure} \[ R^{\theta}\left(\mathbf{e}_{2}\right)=(-\sin (\theta), \cos (\theta)) \overbrace{\cos (\theta)}^{-\sin (\theta)} \overbrace{\mathbf{e}_{1}}^{\mathbf{e}_{2}} x \text { - } \underbrace{\cos (\theta)}_{R^{\theta}\left(\mathbf{e}_{1}\right)}=(\cos (\theta), \sin (\theta)) \] \captionsetup{labelformat=empty} \caption{Figure 1.19: \(R^{\theta}\) rotates the standard basis vectors \(\mathbf{e}_{1}\) and \(\mathbf{e}_{2}\) to \(R^{\theta}\left(\mathbf{e}_{1}\right)=(\cos (\theta), \sin (\theta))\) and \(R^{\theta}\left(\mathbf{e}_{2}\right)=(-\sin (\theta), \cos (\theta))\), respectively.} \end{figure}

\left[R^{\theta}\right]=\left[R^{\theta}\left(\mathbf{e}_{1}\right) \mid R^{\theta}\left(\mathbf{e}_{2}\right)\right]=\left[\begin{array}{cc} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{array}\right] .

\left[R^{\pi / 4}\right]=\left[\begin{array}{cc} \cos (\pi / 4) & -\sin (\pi / 4) \\ \sin (\pi / 4) & \cos (\pi / 4) \end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right] .

\left[R^{-\pi / 6}\right]=\left[\begin{array}{cc} \cos (-\pi / 6) & -\sin (-\pi / 6) \\ \sin (-\pi / 6) & \cos (-\pi / 6) \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc} \sqrt{3} & 1 \\ -1 & \sqrt{3} \end{array}\right] .

\begin{aligned} {\left[R^{\pi / 4}\right] } & =\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right], \quad \text { so } \\ {\left[R^{\pi / 4}\right] \mathbf{v} } & =\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 3 \end{array}\right]=\left[\begin{array}{l} -\sqrt{2} \\ 2 \sqrt{2} \end{array}\right] . \end{aligned}

\left[R^{-\pi / 6}\right] \mathbf{w}=\frac{1}{2}\left[\begin{array}{cc} \sqrt{3} & 1 \\ -1 & \sqrt{3} \end{array}\right]\left[\begin{array}{c} \sqrt{3} \\ 3 \end{array}\right]=\left[\begin{array}{c} 3 \\ \sqrt{3} \end{array}\right] .

\left[R_{y z}^{\theta}\right]=\left[R_{y z}^{\theta}\left(\mathbf{e}_{1}\right)\left|R_{y z}^{\theta}\left(\mathbf{e}_{2}\right)\right| R_{y z}^{\theta}\left(\mathbf{e}_{3}\right)\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos (\theta) & -\sin (\theta) \\ 0 & \sin (\theta) & \cos (\theta) \end{array}\right] .

\left[R_{z x}^{\theta}\right]=\left[\begin{array}{ccc} \cos (\theta) & 0 & -\sin (\theta) \\ 0 & 1 & 0 \\ \sin (\theta) & 0 & \cos (\theta) \end{array}\right] \quad \text { and } \quad\left[R_{x y}^{\theta}\right]=\left[\begin{array}{ccc} \cos (\theta) & -\sin (\theta) & 0 \\ \sin (\theta) & \cos (\theta) & 0 \\ 0 & 0 & 1 \end{array}\right] .

\left[R_{x y}^{2 \pi / 3}\right]=\left[\begin{array}{ccc} \cos (2 \pi / 3) & -\sin (2 \pi / 3) & 0 \\ \sin (2 \pi / 3) & \cos (2 \pi / 3) & 0 \\ 0 & 0 & 1 \end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc} -1 & -\sqrt{3} & 0 \\ \sqrt{3} & -1 & 0 \\ 0 & 0 & 2 \end{array}\right] .

\left[R_{x y}^{2 \pi / 3}\right] \mathbf{v}=\frac{1}{2}\left[\begin{array}{ccc} -1 & -\sqrt{3} & 0 \\ \sqrt{3} & -1 & 0 \\ 0 & 0 & 2 \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \\ 2 \end{array}\right]=\frac{1}{2}\left[\begin{array}{c} \sqrt{3}-3 \\ 3 \sqrt{3}+1 \\ 4 \end{array}\right] \approx\left[\begin{array}{c} -0.6340 \\ 3.0981 \\ 2.000 \end{array}\right] .

(S \circ T)(\mathbf{v})=S(T(\mathbf{v})) \quad \text { for all } \quad \mathbf{v} \in \mathbb{R}^{n} .

[S \circ T]=[S][T] .

(S \circ T)(\mathbf{v})=S(T(\mathbf{v}))=S([T] \mathbf{v})=[S]([T] \mathbf{v})=([S][T]) \mathbf{v} .

\left[F_{\mathbf{u}}\right]=2\left[\begin{array}{l} 3 / 5 \\ 4 / 5 \end{array}\right]\left[\begin{array}{ll} 3 / 5 & 4 / 5 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\frac{1}{25}\left[\begin{array}{cc} -7 & 24 \\ 24 & 7 \end{array}\right] .

[D]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right] .

[T]=[D]\left[F_{\mathbf{u}}\right]=\frac{1}{25}\left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} -7 & 24 \\ 24 & 7 \end{array}\right]=\frac{1}{25}\left[\begin{array}{cc} -14 & 48 \\ 72 & 21 \end{array}\right] .

\sin (\theta+\phi)=\sin (\theta) \cos (\phi)+\cos (\theta) \sin (\phi)

\cos (\theta+\phi)=\cos (\theta) \cos (\phi)-\sin (\theta) \sin (\phi) .

\begin{aligned} & {\left[R^{\theta} \circ R^{\phi}\right]=\left[R^{\theta}\right]\left[R^{\phi}\right]=\left[\begin{array}{cc} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{array}\right]\left[\begin{array}{cc} \cos (\phi) & -\sin (\phi) \\ \sin (\phi) & \cos (\phi) \end{array}\right]} \\ & =\left[\begin{array}{cc} \cos (\theta) \cos (\phi)-\sin (\theta) \sin (\phi) & -\cos (\theta) \sin (\phi)-\sin (\theta) \cos (\phi) \\ \sin (\theta) \cos (\phi)+\cos (\theta) \sin (\phi) & -\sin (\theta) \sin (\phi)+\cos (\theta) \cos (\phi) \end{array}\right] . \end{aligned}

\left[R^{\theta+\phi}\right]=\left[\begin{array}{cc} \cos (\theta+\phi) & -\sin (\theta+\phi) \\ \sin (\theta+\phi) & \cos (\theta+\phi) \end{array}\right] .

\sin (\theta+\phi)=\sin (\theta) \cos (\phi)+\cos (\theta) \sin (\phi),

\cos (\theta+\phi)=\cos (\theta) \cos (\phi)-\sin (\theta) \sin (\phi) .

R=\left[\begin{array}{cc} \cos (1) & -\sin (1) \\ \sin (1) & \cos (1) \end{array}\right] \approx\left[\begin{array}{cc} 0.5403 & -0.8415 \\ 0.8415 & 0.5403 \end{array}\right]

T\left(c_{1} \mathbf{v}_{1}+\cdots+c_{k} \mathbf{v}_{k}\right)=c_{1} T\left(\mathbf{v}_{1}\right)+\cdots+c_{k} T\left(\mathbf{v}_{k}\right)

A=\left[\begin{array}{cc} \cos (\pi / 4) & -\sin (\pi / 4) \\ \sin (\pi / 4) & \cos (\pi / 4) \end{array}\right] .

A=\left[\begin{array}{cc} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{array}\right]

\frac{1}{1+m^{2}}\left[\begin{array}{cc} 1-m^{2} & 2 m \\ 2 m & m^{2}-1 \end{array}\right] .

\left[\begin{array}{cc} \cos (2 \theta) & \sin (2 \theta) \\ \sin (2 \theta) & -\cos (2 \theta) \end{array}\right] .

\left[\begin{array}{ll} 1 & c \\ 0 & 1 \end{array}\right] \quad \text { or } \quad\left[\begin{array}{ll} 1 & 0 \\ c & 1 \end{array}\right]

\begin{aligned} {\left[R_{\mathbf{u}}^{\theta}\right]=} & \cos (\theta) I+(1-\cos (\theta)) \mathbf{u u}^{T} \\ & +\sin (\theta)\left[\begin{array}{ccc} 0 & -u_{3} & u_{2} \\ u_{3} & 0 & -u_{1} \\ -u_{2} & u_{1} & 0 \end{array}\right] . \end{aligned}

[T]=\left[T\left(\mathbf{e}_{1}\right)\left|T\left(\mathbf{e}_{2}\right)\right| \cdots \mid T\left(\mathbf{e}_{n}\right)\right]

A\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 2 x-y \\ y-x \end{array}\right]

A=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right] .

A=\left[\begin{array}{cc} -1 & 2 \\ 0 & 3 \end{array}\right] \quad \text { and } \quad \mathbf{v}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] .

\left[\begin{array}{cc} \cos (\theta) & -\sin (\theta) \\ \sin (\theta) & \cos (\theta) \end{array}\right]^{n}=\left[\begin{array}{cc} \cos (n \theta) & -\sin (n \theta) \\ \sin (n \theta) & \cos (n \theta) \end{array}\right] .

\mathbf{v} \times \mathbf{w} \stackrel{\text { def }}{=}\left(\begin{array}{l} v_{2} w_{3}-v_{3} w_{2} \\ v_{3} w_{1}-v_{1} w_{3} \\ v_{1} w_{2}-v_{2} w_{1} \end{array}\right) .

\begin{aligned} \mathbf{v} \cdot(\mathbf{v} \times \mathbf{w}) & =v_{1}\left(v_{2} w_{3}-v_{3} w_{2}\right)+v_{2}\left(v_{3} w_{1}-v_{1} w_{3}\right)+v_{3}\left(v_{1} w_{2}-v_{2} w_{1}\right) \\ & =v_{1} v_{2} w_{3}-v_{1} v_{3} w_{2}+v_{2} v_{3} w_{1}-v_{2} v_{1} w_{3}+v_{3} v_{1} w_{2}-v_{3} v_{2} w_{1} \\ & =0 . \end{aligned}

\begin{aligned} & \left(v_{1}, v_{2}, v_{3}\right) \times\left(w_{1}, w_{2}, w_{3}\right)=\left(v_{2} w_{3}-v_{3} w_{2}, v_{3} w_{1}-v_{1} w_{3}, v_{1} w_{2}-v_{2} w_{1}\right) \\ & \left(w_{1}, w_{2}, w_{3}\right) \times\left(v_{1}, v_{2}, v_{3}\right)=\left(w_{2} v_{3}-w_{3} v_{2}, w_{3} v_{1}-w_{1} v_{3}, w_{1} v_{2}-w_{2} v_{1}\right) . \end{aligned}

\left(v_{1}, v_{2}, v_{3}\right) \times\left(v_{1}, v_{2}, v_{3}\right)=\left(v_{2} v_{3}-v_{2} v_{3}, v_{1} v_{3}-v_{1} v_{3}, v_{1} v_{2}-v_{1} v_{2}\right)=(0,0,0),

\begin{aligned} \|\mathbf{v} \times \mathbf{w}\|^{2}= & \left\|\left(v_{2} w_{3}-v_{3} w_{2}, v_{3} w_{1}-v_{1} w_{3}, v_{1} w_{2}-v_{2} w_{1}\right)\right\|^{2} \\ = & \left(v_{2} w_{3}-v_{3} w_{2}\right)^{2}+\left(v_{3} w_{1}-v_{1} w_{3}\right)^{2}+\left(v_{1} w_{2}-v_{2} w_{1}\right)^{2} \\ = & v_{2}^{2} w_{3}^{2}+v_{3}^{2} w_{2}^{2}+v_{3}^{2} w_{1}^{2}+v_{1}^{2} w_{3}^{2}+v_{1}^{2} w_{2}^{2}+v_{2}^{2} w_{1}^{2} \\ & -2\left(v_{2} v_{3} w_{2} w_{3}+v_{1} v_{3} w_{1} w_{3}+v_{1} v_{2} w_{1} w_{2}\right) \end{aligned}

\begin{aligned} \|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}-(\mathbf{v} \cdot \mathbf{w})^{2}= & \left(v_{1}^{2}+v_{2}^{2}+v_{3}^{2}\right)\left(w_{1}^{2}+w_{2}^{2}+w_{3}^{2}\right)-\left(v_{1} w_{1}+v_{2} w_{2}+v_{3} w_{3}\right)^{2} \\ = & v_{1}^{2} w_{2}^{2}+v_{1}^{2} w_{3}^{2}+v_{2}^{2} w_{1}^{2}+v_{2}^{2} w_{3}^{2}+v_{3}^{2} w_{1}^{2}+v_{3}^{2} w_{2}^{2} \\ & -2\left(v_{2} v_{3} w_{2} w_{3}+v_{1} v_{3} w_{1} w_{3}+v_{1} v_{2} w_{1} w_{2}\right) \end{aligned}

\begin{aligned} \sqrt{\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}-(\mathbf{v} \cdot \mathbf{w})^{2}} & =\sqrt{\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}-\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2} \cos ^{2}(\theta)} \\ & =\sqrt{\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}\left(1-\cos ^{2}(\theta)\right)} \\ & =\sqrt{\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2} \sin ^{2}(\theta)} \\ & =\|\mathbf{v}\|\|\mathbf{w}\| \sin (\theta) \end{aligned}

\sqrt{\|\mathbf{v}\|^{2}\|\mathbf{w}\|^{2}-(\mathbf{v} \cdot \mathbf{w})^{2}}=\sqrt{10 \cdot 7-(-1)^{2}}=\sqrt{69} .

\begin{aligned} |(1,0,1) \cdot((-1,2,2) \times(3,2,1))| & =|(1,0,1) \cdot(-2,7,-8)| \\ & =|-2-8|=10 . \end{aligned}

\begin{aligned} \mathbf{v} \cdot(\mathbf{w} \times \mathbf{x}) & =-\mathbf{v} \cdot(\mathbf{x} \times \mathbf{w}) \\ & =-\mathbf{w} \cdot(\mathbf{v} \times \mathbf{x}) \\ & =-\mathbf{x} \cdot(\mathbf{w} \times \mathbf{v}) . \end{aligned}

\mathbf{v} \times(\mathbf{w} \times \mathbf{x})+\mathbf{w} \times(\mathbf{x} \times \mathbf{v})+\mathbf{x} \times(\mathbf{v} \times \mathbf{w})=\mathbf{0} .

\mathbf{v} \times \mathbf{w}=\frac{1}{2}(\mathbf{v}-\mathbf{w}) \times(\mathbf{v}+\mathbf{w}) .

A-B-A-D, \quad A-D-A-D, \quad A-D-B-D, \quad \text { and } \quad A-D-C-D .

a_{i, j}=\left\{\begin{array}{l} 1, \text { if there is an edge between the } i-\text { th and } j \text {-th vertices } \\ 0, \text { otherwise. } \end{array}\right.

\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right] .

\left[\begin{array}{lllll} 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}\right] .

\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{array}\right] .

\left[A^{2}\right]_{i, j}=a_{i, 1} a_{1, j}+a_{i, 2} a_{2, j}+\cdots+a_{i, n} a_{n, j} .

A=\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right] .

\begin{aligned} A^{3} & =\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right]\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right]\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right] \\ & =\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{llll} 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right]=\left[\begin{array}{llll} 2 & 3 & 1 & 4 \\ 3 & 2 & 1 & 4 \\ 1 & 1 & 0 & 3 \\ 4 & 4 & 3 & 2 \end{array}\right] . \end{aligned}

A^{6}=\left(A^{3}\right)^{2}=\left[\begin{array}{llll} 2 & 3 & 1 & 4 \\ 3 & 2 & 1 & 4 \\ 1 & 1 & 0 & 3 \\ 4 & 4 & 3 & 2 \end{array}\right]^{2}=\left[\begin{array}{llll} 30 & 29 & 17 & 31 \\ 29 & 30 & 17 & 31 \\ 17 & 17 & 11 & 14 \\ 31 & 31 & 14 & 45 \end{array}\right] .

A=\left[\begin{array}{llllll} 0 & 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 1 & 0 \end{array}\right] .

A^{2}=\left[\begin{array}{llllll} 3 & 2 & 3 & 1 & 2 & 2 \\ 2 & 5 & 3 & 1 & 3 & 3 \\ 3 & 3 & 4 & 1 & 2 & 2 \\ 1 & 1 & 1 & 2 & 2 & 2 \\ 2 & 3 & 2 & 2 & 4 & 3 \\ 2 & 3 & 2 & 2 & 3 & 4 \end{array}\right] .

\left[A^{3}\right]_{4,6}=(0,1,1,0,0,0) \cdot(2,3,2,2,3,4)=3+2=5 .

\left[A^{3}\right]_{4,6}+\left[A^{2}\right]_{4,6}+a_{4,6}=5+2+0=7 .

A=\left[\begin{array}{lllll} 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] .

A^{2}=\left[\begin{array}{lllll} 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] .

\left[A^{4}\right]_{3,5}=(0,0,1,2,1) \cdot(0,0,1,1,0)=0+0+1+2+0=3 .

A=\left[\begin{array}{llll} 0 & 1 & 3 & 0 \\ 1 & 1 & 1 & 1 \\ 3 & 1 & 0 & 2 \\ 0 & 1 & 2 & 0 \end{array}\right] .

\left[A^{2}\right]_{1,4}=(0,1,3,0) \cdot(0,1,2,0)=0+1+6+0=7 .

A=\left[\begin{array}{llll} 0 & 1 & 2 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \end{array}\right] .

A^{2}=\left[\begin{array}{llll} 2 & 1 & 1 & 5 \\ 1 & 2 & 1 & 3 \\ 0 & 3 & 2 & 0 \\ 0 & 1 & 1 & 1 \end{array}\right] .

\left[A^{3}\right]_{2,3}=(1,2,1,3) \cdot(2,1,0,0)=2+2+0+0=4 .

\frac{1}{n} \sum_{j=1}^{n} w_{j} . \quad \begin{aligned} & \text { (average number of friendships }= \\ & \text { total friendships divided by number of people }) \end{aligned}

\begin{array}{lr} \frac{\sum_{j=1}^{n} w_{j}^{2}}{\sum_{j=1}^{n} w_{j}} . & \text { (average number of friends of friends }= \\ \text { total names on lists divided by number of lists) } \end{array}

\sum_{j=1}^{n} w_{j}=|\mathbf{v} \cdot \mathbf{w}| \leq\|\mathbf{v}\|\|\mathbf{w}\|=\sqrt{n} \sqrt{\sum_{j=1}^{n} w_{j}^{2}} .

\frac{1}{n} \sum_{j=1}^{n} w_{j} \leq \frac{\sum_{j=1}^{n} w_{j}^{2}}{\sum_{j=1}^{n} w_{j}},

A=\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right] .

a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=b,

\begin{aligned} x+3 y & =4, & 2 x-\pi y & =3, \\ \sqrt{3} x-y & =\sqrt{5}, & \cos (1) x+\sin (1) y & =2, \end{aligned} \quad \text { and } \quad \begin{aligned} 4 x+3 & =6 y \\ x+y-2 z & =7 . \end{aligned}

\begin{aligned} \sqrt{x}+3 y & =4, & 2 x-7 y^{2} & =3, \\ 2^{x}-2^{y} & =3, & \cos (x)+\sin (y) & =0, \end{aligned} \quad \text { and } \quad \ln (x)-y / z=2 .

\begin{aligned} x+2 y & =4 \\ -x+y & =-1 \end{aligned}

\begin{array}{rlr} x+2 y=4 & x+2 y=4 \\ 2 x+4 y=8 & x+2 y=3 \end{array}

\begin{aligned} a_{1,1} x_{1}+a_{1,2} x_{2}+\cdots+a_{1, n} x_{n} & =b_{1} \\ a_{2,1} x_{1}+a_{2,2} x_{2}+\cdots+a_{2, n} x_{n} & =b_{2} \\ & \vdots \\ a_{m, 1} x_{1}+a_{m, 2} x_{2}+\cdots+a_{m, n} x_{n} & =b_{m} \end{aligned}

\begin{array}{r} x+2 y=4 \\ 3 x+4 y=6 \end{array}

\begin{aligned} & 3 x-2 y+z=-3 \\ & 2 x+3 y-2 z=5 \end{aligned}

\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 4 \\ 6 \end{array}\right]

\left[\begin{array}{ccc} 3 & -2 & 1 \\ 2 & 3 & -2 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -3 \\ 5 \end{array}\right]

\left[\begin{array}{c} 3 x-2 y+z \\ 2 x+3 y-2 z \end{array}\right]=\left[\begin{array}{c} -3 \\ 5 \end{array}\right] .

A\left((1-c) \mathbf{x}_{1}+c \mathbf{x}_{2}\right)=(1-c) A \mathbf{x}_{1}+c A \mathbf{x}_{2}=(1-c) \mathbf{b}+c \mathbf{b}=\mathbf{b}

\begin{array}{r} x+3 y-2 z=5 \\ 2 y-6 z=4 \\ 3 z=6 \end{array}

\begin{aligned} x+3 y & =9 \\ 2 y & =16 \end{aligned}

\begin{aligned} x+3 y-2 z & =5 \\ x+5 y-8 z & =9 \\ 2 x+4 y+5 z & =12 \end{aligned}

\begin{array}{r} (\text { equation } 2) \\ -(\text { equation } 1) \\ \hline(\text { new equation } 2) \end{array}

\begin{aligned} x+5 y-8 z & =9 \\ -x-3 y+2 z & =-5 \\ \hline 2 y-6 z & =4 \end{aligned}

\begin{aligned} \text { (equation 3) } & \begin{aligned} 2 x+4 y+5 z & =12 \\ -2(\text { equation 1 }) & -2 x-6 y+4 z \end{aligned}=-10 \\ \hline \text { (new equation 3) } & -2 y+9 z=2 \end{aligned}

\begin{array}{r} x+3 y-2 z=5 \\ 2 y-6 z=4 \\ -2 y+9 z=2 \end{array}

\begin{array}{r} (\text { equation 3 }) \\ +(\text { equation 2 }) \\ \hline \text { (new equation 3) } \end{array} \quad \begin{array}{r} -2 y+9 z=2 \\ 2 y-6 z=4 \\ \hline 3 z=6 \end{array}

\begin{array}{r} x+3 y-2 z=5 \\ 2 y-6 z=4 \\ 3 z=6 \end{array}

\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 1 & 5 & -8 & 9 \\ 2 & 4 & 5 & 12 \end{array}\right] .

\begin{aligned} x+3 y-2 z & =5 \\ x+5 y-8 z & =9 \\ 2 x+4 y+5 z & =12 \end{aligned}

\begin{gathered} {\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 1 & 5 & -8 & 9 \\ 2 & 4 & 5 & 12 \end{array}\right] \xrightarrow{R_{2}-R_{1}}\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 2 & -6 & 4 \\ 2 & 4 & 5 & 12 \end{array}\right]} \\ \xrightarrow[R_{3}-2 R_{1}]{ }\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 2 & -6 & 4 \\ 0 & -2 & 9 & 2 \end{array}\right] \xrightarrow{R_{3}+R_{2}}\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 2 & -6 & 4 \\ 0 & 0 & 3 & 6 \end{array}\right] . \end{gathered}

\begin{aligned} {\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 2 & -6 & 4 \\ 0 & 0 & 3 & 6 \end{array}\right] \xrightarrow{\frac{1}{3} R_{3}}\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 2 & -6 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right] } \\ \xrightarrow{\substack{R_{1}+2 R_{3} \\ R_{2}+6 R_{3}}}\left[\begin{array}{lll|c} 1 & 3 & 0 & 9 \\ 0 & 2 & 0 & 16 \\ 0 & 0 & 1 & 2 \end{array}\right] \xrightarrow{\frac{1}{2} R_{2}}\left[\begin{array}{lll|c} 1 & 3 & 0 & 9 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{array}\right] \\ \xrightarrow{R_{1}-3 R_{2}}\left[\begin{array}{lll|c} 1 & 0 & 0 & -15 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{array}\right] . \end{aligned}

\left[\begin{array}{ll|l} 1 & 2 & 3 \\ 3 & 4 & 7 \end{array}\right] \xrightarrow{0 R_{2}}\left[\begin{array}{ll|l} 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}\right],

\left[\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 2 & -6 & 4 \\ 0 & 0 & 3 & 6 \end{array}\right] \quad \text { and } \quad\left[\begin{array}{ccc|c} 1 & 0 & 0 & -15 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{array}\right]

\left[\begin{array}{llll} \star & * & * & * \\ 0 & 0 & \star & * \\ 0 & 0 & 0 & \star \\ 0 & 0 & 0 & 0 \end{array}\right] \quad \text { and } \quad\left[\begin{array}{llllllllll} 0 & 0 & \star & * & * & * & * & * & * & * \\ 0 & 0 & 0 & \star & * & * & * & * & * & * \\ 0 & 0 & 0 & 0 & 0 & \star & * & * & * & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \star & * \end{array}\right] .

\left[\begin{array}{llll} 1 & * & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \quad \text { and } \quad\left[\begin{array}{llllllllll} 0 & 0 & 1 & 0 & * & 0 & * & * & 0 & * \\ 0 & 0 & 0 & 1 & * & 0 & * & * & 0 & * \\ 0 & 0 & 0 & 0 & 0 & 1 & * & * & 0 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & * \end{array}\right] .

\left[\begin{array}{ccccc} 0 & 0 & -1 & 1 & 0 \\ 0 & -2 & 1 & -5 & 2 \\ 0 & 2 & -2 & 6 & -3 \\ 0 & -4 & 2 & -10 & 5 \end{array}\right] .

\left[\begin{array}{ccccc} 0 & 0 & -1 & 1 & 0 \\ 0 & -2 & 1 & -5 & 2 \\ 0 & 2 & -2 & 6 & -3 \\ 0 & -4 & 2 & -10 & 5 \end{array}\right] \xrightarrow{R_{1} \leftrightarrow R_{3}}\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & -3 \\ 0 & -2 & 1 & -5 & 2 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & -4 & 2 & -10 & 5 \end{array}\right]

\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & -3 \\ 0 & -2 & 1 & -5 & 2 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & -4 & 2 & -10 & 5 \end{array}\right] \xrightarrow{\substack{R_{2}+R_{1} \\ R_{4}+2 R_{1}}}\left[\begin{array}{cccc} 0 & 2 & & \begin{array}{ccc} -2 & 6 & -3 \\ 0 & 0 & \leftarrow \\ 0 & 0 & 1 \end{array} \\ 0 & 0 & -1 \\ -1 & 1 & 0 \\ -2 & 2 & -1 \end{array}\right]

\rightarrow\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & -3 \\ \hline 0 & 0 & -1 & 1 & -1 \\ 0 & 0 & \uparrow-1 & 1 & 0 \\ 0 & 0 & -2 & 2 & -1 \end{array}\right] \xrightarrow{\substack{R_{3}-R_{2} \\ R_{4}-2 R_{2}}}\left[\begin{array}{ccc|cc} 0 & 2 & -2 & 6 & -3 \\ \hline 0 & 0 & -1 & 1 & -1 \\ 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 \end{array}\right]

\begin{gathered} {\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & -3 \\ 0 & 0 & -1 & 1 & -1 \\ \hline 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & \uparrow \end{array}\right] \xrightarrow{R_{4}-R_{3}}\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & -3 \\ 0 & 0 & -1 & 1 & -1 \\ \hline 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]} \\ \text { new leading entry } \end{gathered}

\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & -3 \\ 0 & 0 & -1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_{1}+3 R_{3} \\ R_{2}+R_{3}}}\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \text { new zeros }

\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & 0 \\ 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{-R_{2}}\left[\begin{array}{ccccc} 0 & 2 & -2 & 6 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \quad \begin{gathered} \text { new zero } \\ \downarrow \end{gathered}

\xrightarrow{R_{1}+2 R_{2}}\left[\begin{array}{ccccc} 0 & 2 & 0 & 4 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

\text { next leading entry }\left[\begin{array}{ccccc} 0 & 2 & 0 & 4 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{\frac{1}{2} R_{1}}\left[\begin{array}{ccccc} 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

\left[\begin{array}{cccc|c} 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

\begin{aligned} x+2 z & =0 \\ y-z & =0 \\ 0 & =1 \\ 0 & =0 \end{aligned}

\left.\left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 2 \end{array}\right] \quad \right\rvert\, \quad \begin{aligned} & x=4 \\ & y=-3 \\ & z=2 \end{aligned}

\left[\begin{array}{lll|l} 1 & 4 & 0 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]

\mid

\begin{aligned} x+4 y & =3 \\ z & =2 \\ 0 & =0 \end{aligned}

\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 3-4 y \\ y \\ 2 \end{array}\right)=\left(\begin{array}{l} 3 \\ 0 \\ 2 \end{array}\right)+y\left(\begin{array}{c} -4 \\ 1 \\ 0 \end{array}\right) .

\left[\begin{array}{ccc|c} 1 & 2 & -2 & -4 \\ 2 & 4 & 1 & 0 \\ 1 & 2 & 7 & 2 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{ccc|c} 1 & 2 & -2 & -4 \\ 2 & 4 & 1 & 0 \\ 1 & 2 & 7 & 2 \end{array}\right] } & \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{ccc|c} 1 & 2 & -2 & -4 \\ 0 & 0 & 5 & 8 \\ 0 & 0 & 9 & 6 \end{array}\right] \\ & \xrightarrow{R_{3}-\frac{9}{5} R_{2}}\left[\begin{array}{ccc|c} 1 & 2 & -2 & -4 \\ 0 & 0 & 5 & 8 \\ 0 & 0 & 0 & -42 / 5 \end{array}\right] \end{aligned}

\begin{aligned} v_{1}+2 v_{2}+3 v_{3} & =0 \\ v_{2}-v_{3} & =0 \end{aligned}

\begin{aligned} v_{1}+2 v_{2}+2 v_{3}+2 v_{4} & =0 \\ 2 v_{1}+v_{2}-v_{3} & =0 \\ v_{1}+2 v_{3}+v_{4} & =0 \end{aligned}

\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 / 2 & 0 \\ 0 & 0 & 1 & 1 / 2 & 0 \end{array}\right] .

(1,1,3)=c_{1}(-1,3,2)+c_{2}(3,1,-1) .

\begin{aligned} -c_{1}+3 c_{2} & =1 \\ 3 c_{1}+c_{2} & =1 \\ 2 c_{1}-c_{2} & =3 \end{aligned}

\begin{aligned} {\left[\begin{array}{cc|c} -1 & 3 & 1 \\ 3 & 1 & 1 \\ 2 & -1 & 3 \end{array}\right] } & \xrightarrow{R_{2}+3 R_{1}}\left[\begin{array}{cc|c} -1 & 3 & 1 \\ R_{3}+2 R_{1} & 10 & 4 \\ 0 & 5 & 5 \end{array}\right] \\ & \xrightarrow{R_{3}-\frac{1}{2} R_{2}}\left[\begin{array}{cc|c} -1 & 3 & 1 \\ 0 & 10 & 4 \\ 0 & 0 & 3 \end{array}\right] . \end{aligned}

(1,3,1)=c_{1}(-1,3,2)+c_{2}(3,1,-1) .

\begin{aligned} {\left[\begin{array}{cc|c} -1 & 3 & 1 \\ 3 & 1 & 3 \\ 2 & -1 & 1 \end{array}\right] } & \xrightarrow{R_{2}+3 R_{1}}\left[\begin{array}{cc|c} -1 & 3 & 1 \\ R_{3}+2 R_{1} & 10 & 6 \\ 0 & 5 & 3 \end{array}\right] \\ & \xrightarrow{R_{3}-\frac{1}{2} R_{2}}\left[\begin{array}{cc|c} -1 & 3 & 1 \\ 0 & 10 & 6 \\ 0 & 0 & 0 \end{array}\right] . \end{aligned}

(1,3,1)=\frac{4}{5}(-1,3,2)+\frac{3}{5}(3,1,-1) .

x+y=500 .

0.035 x+0.01 y=10 .

\begin{aligned} & {\left[\begin{array}{cc|c} 1 & 1 & 500 \\ 0.035 & 0.01 & 10 \end{array}\right] \xrightarrow{R_{2}-0.035 R_{1}}\left[\begin{array}{cc|c} 1 & 1 & 500 \\ 0 & -0.025 & -7.5 \end{array}\right]} \\ & \quad \xrightarrow{-40 R_{2}}\left[\begin{array}{ll|l} 1 & 1 & 500 \\ 0 & 1 & 300 \end{array}\right] \xrightarrow{R_{1}-R_{2}}\left[\begin{array}{ll|l} 1 & 0 & 200 \\ 0 & 1 & 300 \end{array}\right] . \end{aligned}

\mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} .

w \mathrm{C}_{4} \mathrm{H}_{10}+x \mathrm{O}_{2} \rightarrow y \mathrm{CO}_{2}+z \mathrm{H}_{2} \mathrm{O}

2 x=2 y+z, \quad \text { or equivalently } \quad 2 x-2 y-z=0 .

4 w-y=0 \quad \text { and } \quad 10 w-2 z=0 .

\left[\begin{array}{cccc|c} 0 & 2 & -2 & -1 & 0 \\ 4 & 0 & -1 & 0 & 0 \\ 10 & 0 & 0 & -2 & 0 \end{array}\right] \xrightarrow{\text { row-reduce }}\left[\begin{array}{cccc|c} 1 & 0 & 0 & -1 / 5 & 0 \\ 0 & 1 & 0 & -13 / 10 & 0 \\ 0 & 0 & 1 & -4 / 5 & 0 \end{array}\right] .

2 \mathrm{C}_{4} \mathrm{H}_{10}+13 \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O} .

\left[\begin{array}{cccc} 3 & -2 & 0 & -2 \\ 0 & -2 & -2 & 2 \\ 0 & 2 & 0 & -1 \\ 1 & 0 & 2 & -2 \end{array}\right]

\left[\begin{array}{ccccc} 4 & 2 & -1 & 2 & 1 \\ 1 & 2 & -1 & 0 & 4 \\ 5 & 1 & 2 & 6 & -1 \\ -3 & 4 & 2 & 2 & 5 \end{array}\right]

\left[\begin{array}{cccccc} 0 & -2 & -2 & 3 & 4 & 1 \\ 1 & 0 & -2 & 5 & 4 & 0 \\ 5 & 0 & 2 & 1 & 0 & 3 \\ -1 & 1 & 0 & 2 & 4 & -1 \end{array}\right]

\left[\begin{array}{ccccccc} 1 & -1 & -3 & -1 & -3 & -3 & -1 \\ 0 & -1 & -2 & 1 & -2 & 3 & 2 \\ 3 & -1 & -5 & 0 & 0 & -5 & 3 \\ 2 & -3 & -8 & -2 & -9 & -5 & 1 \\ -2 & 3 & 8 & -2 & 5 & -3 & -1 \end{array}\right]

\begin{aligned} 2 y+3 z & =-1 \\ z & =1 \end{aligned}

\begin{array}{r} x+2 y=3 \\ 2 x+y=3 \end{array}

\begin{aligned} & x+y+z=4 \\ & x-y+z=0 \end{aligned}

\begin{aligned} x-y & =2 \\ x+2 y & =4 \\ 2 x-y & =5 \end{aligned}

\begin{aligned} 2 x+y-z & =1 \\ x-3 y+z & =-2 \\ 2 x-2 y+3 z & =7 \end{aligned}

\begin{aligned} x+y+z & =1 \\ -x+z & =2 \\ 2 x+y & =0 \end{aligned}

\begin{aligned} w+x+y-z & =0 \\ 2 x+3 y+z & =-1 \\ y-z & =-3 \\ 3 z & =3 \end{aligned}

\begin{aligned} v-2 w-x-2 y-z & =1 \\ 2 v-2 x-6 y-4 z & =2 \\ 4 w+y-2 z & =3 \end{aligned}

\begin{aligned} & 6 v+5 w+3 x-2 y-2 z=1 \\ & 3 v-w+x+5 y+4 z=2 \\ & 3 v+4 w+x+3 y+4 z=3 \\ & 2 v+6 w+x-2 y-z=4 \end{aligned}

\begin{aligned} v-w+2 x+6 y+6 z & =3 \\ 4 v+3 w-y+4 z & =0 \\ 5 v+w-2 x-2 y-2 z & =2 \\ w-2 x+3 y-2 z & =-1 \\ v+5 w-x+5 z & =3 \end{aligned}

\left[\begin{array}{cc|c} 1 & -1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & t \end{array}\right] .

\begin{array}{r} x+y+h z=1 \\ y-z=k \\ x-y+2 z=3 \end{array}

\begin{array}{r} w+x / 2+y / 3+z / 4=1 \\ w / 2+x / 3+y / 4+z / 5=1 \\ w / 3+x / 4+y / 5+z / 6=1 \\ w / 4+x / 5+y / 6+z / 7=h \end{array}

\left[\begin{array}{ll|l} a & b & 1 \\ c & d & 1 \end{array}\right] .

\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]

\left[\begin{array}{ll|l} 1 & 1 & 3 \\ 1 & 2 & 5 \end{array}\right] .

\left(v_{2} w_{3}-v_{3} w_{2}, v_{3} w_{1}-v_{1} w_{3}, v_{1} w_{2}-v_{2} w_{1}\right)

\mathrm{ZnS}+\mathrm{O}_{2} \rightarrow \mathrm{ZnO}+\mathrm{SO}_{2} .

\mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} .

\begin{aligned} 1 / x+2 / y & =3 / x y \\ x+y & =2 \end{aligned}

\begin{aligned} y / x+2 x / y & =6 / x y \\ x^{2}+y^{2} & =5 \end{aligned}

\begin{aligned} 1 / x-2 / z & =-4 \\ 1 / x+1 / y+1 / z & =4 \\ 2 / x-6 / y-2 / z & =4 \end{aligned}

\begin{aligned} \sin (x)+2 \cos (y)-\cos (z) & =3 \\ 2 \sin (x)+\cos (y)+\cos (z) & =0 \\ -\sin (x)-\cos (y)+\cos (z) & =-2 \end{aligned}

A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] .

B=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]

C=\left[\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right] .

p(x)=a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f

\left[\begin{array}{cccc} 0 & 2 & 4 & 0 \\ 1 & 1 & 0 & -1 \\ 3 & 4 & 2 & 1 \end{array}\right] .

\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} 0 & 2 & 4 & 0 \\ 1 & 1 & 0 & -1 \\ 3 & 4 & 2 & 1 \end{array}\right]=\left[\begin{array}{cccc} 1 & 1 & 0 & -1 \\ 0 & 2 & 4 & 0 \\ 3 & 4 & 2 & 1 \end{array}\right] .

\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} 1 & 1 & 0 & -1 \\ 0 & 2 & 4 & 0 \\ 3 & 4 & 2 & 1 \end{array}\right]=\left[\begin{array}{cccc} 1 & 1 & 0 & -1 \\ 0 & 2 & 4 & 0 \\ 0 & 1 & 2 & 4 \end{array}\right],

\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 / 2 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} 1 & 1 & 0 & -1 \\ 0 & 2 & 4 & 0 \\ 0 & 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{cccc} 1 & 1 & 0 & -1 \\ 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & 4 \end{array}\right] .

\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right],

\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right] \quad \text { and } \quad\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 / 2 & 0 \\ 0 & 0 & 1 \end{array}\right],

A=\left[\begin{array}{cccc} 0 & 2 & 4 & 0 \\ 1 & 1 & 0 & -1 \\ 3 & 4 & 2 & 1 \end{array}\right] .

E_{1}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right], \quad E_{2}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right], \quad \text { and } \quad E_{3}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 / 2 & 0 \\ 0 & 0 & 1 \end{array}\right],

E_{3} E_{2} E_{1} A=\left[\begin{array}{cccc} 1 & 1 & 0 & -1 \\ 0 & 1 & 2 & 0 \\ 0 & 1 & 2 & 4 \end{array}\right],

E_{4}=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \quad \text { and } \quad E_{5}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right],

E_{5} E_{4} E_{3} E_{2} E_{1} A=\left[\begin{array}{cccc} 1 & 0 & -2 & -1 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 4 \end{array}\right] .

E_{6}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 / 4 \end{array}\right] \quad \text { and } \quad E_{7}=\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right],

E_{7} E_{6} E_{5} E_{4} E_{3} E_{2} E_{1} A=\left[\begin{array}{cccc} 1 & 0 & -2 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right],

E=E_{7} E_{6} E_{5} E_{4} E_{3} E_{2} E_{1}=\frac{1}{8}\left[\begin{array}{ccc} -5 & 2 & 2 \\ 4 & 0 & 0 \\ -1 & -6 & 2 \end{array}\right] .

A=\left[\begin{array}{cccc} 0 & 2 & 4 & 0 \\ 1 & 1 & 0 & -1 \\ 3 & 4 & 2 & 1 \end{array}\right] .

\begin{aligned} & {\left[\begin{array}{cccc|ccc} 0 & 2 & 4 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & -1 & 0 & 1 & 0 \\ 3 & 4 & 2 & 1 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{1} \leftrightarrow R_{2}}\left[\begin{array}{cccc|ccc} 1 & 1 & 0 & -1 & 0 & 1 & 0 \\ 0 & 2 & 4 & 0 & 1 & 0 & 0 \\ 3 & 4 & 2 & 1 & 0 & 0 & 1 \end{array}\right]} \\ & \xrightarrow{R_{3}-3 R_{1}}\left[\begin{array}{cccc|ccc} 1 & 1 & 0 & -1 & 0 & 1 & 0 \\ 0 & 2 & 4 & 0 & 1 & 0 & 0 \\ 0 & 1 & 2 & 4 & 0 & -3 & 1 \end{array}\right] \\ & \xrightarrow{\frac{1}{2} R_{2}}\left[\begin{array}{cccc|ccc} 1 & 1 & 0 & -1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 & 1 / 2 & 0 & 0 \\ 0 & 1 & 2 & 4 & 0 & -3 & 1 \end{array}\right] \\ & \xrightarrow{R_{1}-R_{2}}\left[\begin{array}{cccc|ccc} 1 & 0 & -2 & -1 & -1 / 2 & 1 & 0 \\ R_{3}-R_{2} & 1 & 2 & 0 & 1 / 2 & 0 & 0 \\ 0 & 0 & 0 & 4 & -1 / 2 & -3 & 1 \end{array}\right] . \end{aligned}

\begin{gathered} {\left[\begin{array}{cccc|ccc} 1 & 0 & -2 & -1 & -1 / 2 & 1 & 0 \\ 0 & 1 & 2 & 0 & 1 / 2 & 0 & 0 \\ 0 & 0 & 0 & 4 & -1 / 2 & -3 & 1 \end{array}\right]} \\ \xrightarrow{\frac{1}{4} R_{3}}\left[\begin{array}{cccc|ccc} 1 & 0 & -2 & -1 & -1 / 2 & 1 & 0 \\ 0 & 1 & 2 & 0 & 1 / 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 / 8 & -3 / 4 & 1 / 4 \end{array}\right] \\ \xrightarrow{R_{1}+R_{3}}\left[\begin{array}{cccc|ccc} 1 & 0 & -2 & 0 & -5 / 8 & 1 / 4 & 1 / 4 \\ 0 & 1 & 2 & 0 & 1 / 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 / 8 & -3 / 4 & 1 / 4 \end{array}\right] . \end{gathered}

[R \mid E]=E_{k} \cdots E_{2} E_{1}[A \mid I]=\left[E_{k} \cdots E_{2} E_{1} A \mid E_{k} \cdots E_{2} E_{1}\right] .

E_{1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right] \quad \text { and } \quad E_{2}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{array}\right],

A A^{-1}=A^{-1} A=I .

B=I B=(C A) B=C(A B)=C I=C,

\begin{aligned} & {\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left(\frac{1}{2}\left[\begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array}\right]\right)=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad \text { and }} \\ & \left(\frac{1}{2}\left[\begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array}\right]\right)\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] . \end{aligned}

\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right]

\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} -4 & 2 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] .

\begin{array}{lll} {\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]\left(\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\right)} & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] & \text { and } \\ \left(\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\right)\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] & =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] . \end{array}

(c A)\left(\frac{1}{c} A^{-1}\right)=\frac{c}{c} A A^{-1}=I \quad \text { and } \quad\left(\frac{1}{c} A^{-1}\right)(c A)=\frac{c}{c} A^{-1} A=I

A^{-k} \stackrel{\text { def }}{=}\left(A^{-1}\right)^{k} \quad \text { for all integers } \quad k \geq 1,

A^{-1}=\frac{1}{2}\left[\begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array}\right] .

A^{-2}=\left(A^{-1}\right)^{2}=\left(\frac{1}{2}\left[\begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array}\right]\right)\left(\frac{1}{2}\left[\begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array}\right]\right)=\frac{1}{4}\left[\begin{array}{cc} 22 & -10 \\ -15 & 7 \end{array}\right] .

\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 / 3 \end{array}\right] .

\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right] .

\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] .

\left[\begin{array}{lll} 1 & 0 & 0 \\ 5 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] .

(A B)^{-1}=B^{-1} A^{-1} .

\begin{aligned} & (A B)\left(B^{-1} A^{-1}\right)=A\left(B B^{-1}\right) A^{-1}=A I A^{-1}=A A^{-1}=I, \quad \text { and } \\ & \left(B^{-1} A^{-1}\right)(A B)=B^{-1}\left(A^{-1} A\right) B=B^{-1} I B=B^{-1} B=I . \end{aligned}

\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right] \quad \text { and } \quad\left[\begin{array}{ll} 1 & 0 \\ 0 & 3 \end{array}\right]^{-1}=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 / 3 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{ll} 1 & 6 \\ 0 & 3 \end{array}\right]^{-1}=\left(\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 3 \end{array}\right]\right)^{-1} } & =\left[\begin{array}{ll} 1 & 0 \\ 0 & 3 \end{array}\right]^{-1}\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]^{-1} \\ & =\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 / 3 \end{array}\right]\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 / 3 \end{array}\right] . \end{aligned}

(A B C)^{-1}=((A B) C)^{-1}=C^{-1}(A B)^{-1}=C^{-1} B^{-1} A^{-1} .

A\left(A^{-1} \mathbf{b}\right)=\left(A A^{-1}\right) \mathbf{b}=I \mathbf{b}=\mathbf{b} .

\mathbf{x}-\mathbf{y}=\left(A^{-1} A\right)(\mathbf{x}-\mathbf{y})=A^{-1}(A(\mathbf{x}-\mathbf{y}))=A^{-1} \mathbf{0}=\mathbf{0},

A=E_{1}^{-1} E_{2}^{-1} \cdots E_{k}^{-1} .

\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right] \xrightarrow{R_{2}-2 R_{1}}\left[\begin{array}{ll} 1 & 2 \\ 0 & 0 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{ccc} 1 & 2 & -2 \\ 0 & 1 & -2 \\ 1 & 1 & 1 \end{array}\right] } & \xrightarrow{R_{3}-R_{1}}\left[\begin{array}{ccc} 1 & 2 & -2 \\ 0 & 1 & -2 \\ 0 & -1 & 3 \end{array}\right] \\ & \xrightarrow{\substack{R_{1}-2 R_{2} \\ R_{3}+R_{2}}}\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right] \xrightarrow{\substack{R_{1}-2 R_{3} \\ R_{2}+2 R_{3}}}\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] . \end{aligned}

\begin{aligned} {\left[\begin{array}{ll|ll} 2 & 2 & 1 & 0 \\ 4 & 5 & 0 & 1 \end{array}\right] \xrightarrow{R_{2}-2 R_{1}}\left[\begin{array}{ll|cc} 2 & 2 & 1 & 0 \\ 0 & 1 & -2 & 1 \end{array}\right] } \\ \quad \xrightarrow{R_{1}-2 R_{2}}\left[\begin{array}{ll|cc} 2 & 0 & 5 & -2 \\ 0 & 1 & -2 & 1 \end{array}\right] \xrightarrow{\frac{1}{2} R_{1}}\left[\begin{array}{cc|cc} 1 & 0 & 5 / 2 & -1 \\ 0 & 1 & -2 & 1 \end{array}\right] . \end{aligned}

\left[\begin{array}{cc} 5 / 2 & -1 \\ -2 & 1 \end{array}\right]

\left[\begin{array}{cc|cc} 1 & -2 & 1 & 0 \\ -3 & 6 & 0 & 1 \end{array}\right] \xrightarrow{R_{2}+3 R_{1}}\left[\begin{array}{cc|cc} 1 & -2 & 1 & 0 \\ 0 & 0 & 3 & 1 \end{array}\right] .

\begin{gathered} {\left[\begin{array}{lll|lll} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}\right] \xrightarrow{\substack{R_{2}-4 R_{1} \\ R_{3}-7 R_{1}}}\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & -6 & -12 & -7 & 0 & 1 \end{array}\right]} \\ \xrightarrow{R_{3}-2 R_{2}}\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{array}\right] . \end{gathered}

\begin{array}{r} {\left[\begin{array}{lll|lll} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 2 & 4 & 0 & 1 & 0 \\ 1 & 3 & 9 & 0 & 0 & 1 \end{array}\right] \xrightarrow{\substack{R_{2}-R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{lll|ccl} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 3 & -1 & 1 & 0 \\ 0 & 2 & 8 & -1 & 0 & 1 \end{array}\right]} \\ \xrightarrow{R_{1}-R_{2}}\left[\begin{array}{ccc|ccc} 1 & 0 & -2 & 2 & -1 & 0 \\ 0 & 1 & 3 & -1 & 1 & 0 \\ 0 & 0 & 2 & 1 & -2 & 1 \end{array}\right] \\ \xrightarrow{\frac{1}{2} R_{3}}\left[\begin{array}{ccc|ccc} 1 & 0 & -2 & 2 & -1 & 0 \\ 0 & 1 & 3 & -1 & 1 & 0 \\ 0 & 0 & 1 & 1 / 2 & -1 & 1 / 2 \end{array}\right] \\ \xrightarrow{R_{1}+2 R_{3}}\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & -3 & 1 \\ 0 & 1 & 0 & -5 / 2 & 4 & -3 / 2 \\ 0 & 0 & 1 & 1 / 2 & -1 & 1 / 2 \end{array}\right] . \end{array}

\left[\begin{array}{ccc} 3 & -3 & 1 \\ -5 / 2 & 4 & -3 / 2 \\ 1 / 2 & -1 & 1 / 2 \end{array}\right]

A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] .

\left(\frac{1}{a d-b c}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]\right)\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\frac{1}{a d-b c}\left[\begin{array}{cc} a d-b c & 0 \\ 0 & a d-b c \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right],

\frac{1}{-2}\left[\begin{array}{cc} 4 & -2 \\ -3 & 1 \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array}\right] .

\frac{1}{2}\left[\begin{array}{cc} 5 & -4 \\ -2 & 2 \end{array}\right] .

A \mathbf{x}=A\left(A^{-1} \mathbf{b}\right)=\left(A^{-1} A\right) \mathbf{b}=I \mathbf{b}=\mathbf{b} .

A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{array}\right] \quad \text { and } \quad \mathbf{b}=\left[\begin{array}{l} 2 \\ 1 \\ 2 \end{array}\right] .

A^{-1}=\left[\begin{array}{ccc} 3 & -3 & 1 \\ -5 / 2 & 4 & -3 / 2 \\ 1 / 2 & -1 & 1 / 2 \end{array}\right] .

\mathbf{x}=A^{-1} \mathbf{b}=\left[\begin{array}{ccc} 3 & -3 & 1 \\ -5 / 2 & 4 & -3 / 2 \\ 1 / 2 & -1 & 1 / 2 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 2 \end{array}\right]=\left[\begin{array}{c} 5 \\ -4 \\ 1 \end{array}\right] .

\mathbf{b}=\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \quad \text { then } \quad \mathbf{x}=A^{-1} \mathbf{b}=\left[\begin{array}{ccc} 3 & -3 & 1 \\ -5 / 2 & 4 & -3 / 2 \\ 1 / 2 & -1 & 1 / 2 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\left[\begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right] .

A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{array}\right] .

B A=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 / 2 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] .

\left[\begin{array}{lll} a & 1 & 1 \\ 0 & a & 1 \\ 0 & 0 & a \end{array}\right] .

A=\left[\begin{array}{ccccc} a & b & b & \cdots & b \\ b & a & b & \cdots & b \\ b & b & a & \cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \cdots & a \end{array}\right] \in \mathcal{M}_{n} .

A=\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] .

\left[\begin{array}{ll} A & O \\ O & D \end{array}\right]^{-1}=\left[\begin{array}{cc} A^{-1} & O \\ O & D^{-1} \end{array}\right],

\left[\begin{array}{cccc} A_{1} & O & \cdots & O \\ O & A_{2} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & A_{n} \end{array}\right]

\begin{aligned} & {\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]^{-1}=} \\ & {\left[\begin{array}{cc} A^{-1}+A^{-1} B S^{-1} C A^{-1} & -A^{-1} B S^{-1} \\ -S^{-1} C A^{-1} & S^{-1} \end{array}\right] .} \end{aligned}

\left(A+\mathbf{v} \mathbf{w}^{T}\right)^{-1}=A^{-1}-\frac{A^{-1} \mathbf{v w}^{T} A^{-1}}{1+\mathbf{w}^{T} A^{-1} \mathbf{v}}

A_{n}=\left[\begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 1 & \cdots & 1 \\ 1 & 1 & 3 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & n \end{array}\right] .

\mathbf{0}=0 \mathbf{v}=c \mathbf{v} \in \mathcal{S} .

c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k} \in \mathcal{S}

\left(v_{1}+w_{1}\right)+\left(v_{2}+w_{2}\right)-3\left(v_{3}+w_{3}\right)=0,

\left(c v_{1}\right)+\left(c v_{2}\right)-3\left(c v_{3}\right)=0,

A(\mathbf{v}+\mathbf{w})=A \mathbf{v}+A \mathbf{w}=\mathbf{0}+\mathbf{0}=\mathbf{0}

A(c \mathbf{v})=c(A \mathbf{v})=c \mathbf{0}=\mathbf{0}

A \mathbf{x}=\left[\begin{array}{ll} 2 & -2 \\ 1 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 2(x-y) \\ x-y \end{array}\right] .

\left[\begin{array}{ll|l} 2 & -2 & 0 \\ 1 & -1 & 0 \end{array}\right] \xrightarrow{R_{1} \leftrightarrow R_{2}}\left[\begin{array}{ll|l} 1 & -1 & 0 \\ 2 & -2 & 0 \end{array}\right] \xrightarrow{R_{2}-2 R_{1}}\left[\begin{array}{cc|c} 1 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right],

\operatorname{span}(B) \quad \text { or } \quad \operatorname{span}\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right) .

(x, y)=x \mathbf{e}_{1}+y \mathbf{e}_{2} .

\mathbf{v}=\left(v_{1}, v_{2}, \ldots, v_{n}\right)=v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n} .

(x, y)=c_{1}(1,2)+c_{2}(2,1) .

\begin{array}{r} c_{1}+2 c_{2}=x \\ 2 c_{1}+c_{2}=y \end{array}

\left[\begin{array}{ll|l} 1 & 2 & x \\ 2 & 1 & y \end{array}\right] \xrightarrow{R_{2}-2 R_{1}}\left[\begin{array}{cc|c} 1 & 2 & x \\ 0 & -3 & y-2 x \end{array}\right] .

(x, y, z)=c_{1}(1,2,1)+c_{2}(2,1,1) .

\begin{aligned} {\left[\begin{array}{ll|l} 1 & 2 & x \\ 2 & 1 & y \\ 1 & 1 & z \end{array}\right] } & \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{cc|c} 1 & 2 & x \\ 0 & -3 & y-2 x \\ 0 & -1 & z-x \end{array}\right] \\ & \xrightarrow{R_{3}-\frac{1}{3} R_{2}}\left[\begin{array}{cc|c} 1 & 2 & x \\ 0 & -3 & y-2 x \\ 0 & 0 & z-\frac{1}{3} x-\frac{1}{3} y \end{array}\right] . \end{aligned}

\begin{aligned} \mathbf{v} & =c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k} \\ \mathbf{w} & =d_{1} \mathbf{v}_{1}+d_{2} \mathbf{v}_{2}+\cdots+d_{k} \mathbf{v}_{k} . \end{aligned}

\mathbf{v}+\mathbf{w}=\left(c_{1}+d_{1}\right) \mathbf{v}_{1}+\left(c_{2}+d_{2}\right) \mathbf{v}_{2}+\cdots+\left(c_{k}+d_{k}\right) \mathbf{v}_{k},

c \mathbf{v}=\left(c c_{1}\right) \mathbf{v}_{1}+\left(c c_{2}\right) \mathbf{v}_{2}+\cdots+\left(c c_{k}\right) \mathbf{v}_{k},

\operatorname{range}(A)=\operatorname{span}\left(\mathbf{a}_{1}, \mathbf{a}_{2}, \ldots, \mathbf{a}_{n}\right) .

A \mathbf{x}=x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+\cdots+x_{n} \mathbf{a}_{n} .

\operatorname{range}(A)=\operatorname{span}((2,1),(-2,-1)) .

\operatorname{range}(A)=\operatorname{span}((2,1)) .

\left[\begin{array}{cc|c} 1 & -1 & 2 \\ -1 & 1 & -2 \end{array}\right] \quad \text { is } \quad\left[\begin{array}{cc|c} 1 & -1 & 2 \\ 0 & 0 & 0 \end{array}\right] .

c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0} .

(2,3)-2(1,0)-3(0,1)=(0,0) .

0(1,0,0)+0(0,1,0)+0(0,0,1)=(0,0,0) .

c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0}

c_{1}(1,-1,0)+c_{2}(-2,1,2)+c_{3}(1,1,-4)=(0,0,0) .

\begin{aligned} c_{1}-2 c_{2}+c_{3} & =0 \\ -c_{1}+c_{2}+c_{3} & =0 \\ 2 c_{2}-4 c_{3} & =0 \end{aligned}

\begin{aligned} {\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ -1 & 1 & 1 & 0 \\ 0 & 2 & -4 & 0 \end{array}\right] \xrightarrow{R_{2}+R_{1}} } & {\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -4 & 0 \end{array}\right] } \\ & \xrightarrow{R_{3}+2 R_{2}}\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

3(1,-1,0)+2(-2,1,2)+(1,1,-4)=(0,0,0) .

c_{1}(1,2,3)+c_{2}(1,0,1)+c_{3}(0,-1,2)=(0,0,0) .

\begin{aligned} {\left[\begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 2 & 0 & -1 & 0 \\ 3 & 1 & 2 & 0 \end{array}\right] } & \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-3 R_{1}}}\left[\begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & -2 & -1 & 0 \\ 0 & -2 & 2 & 0 \end{array}\right] \\ & \xrightarrow{R_{3}-R_{2}}\left[\begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & -2 & -1 & 0 \\ 0 & 0 & 3 & 0 \end{array}\right] \end{aligned}

A \mathbf{x}=\left[\begin{array}{l|l|ll} \mathbf{a}_{1} \mid & \mathbf{a}_{2} \mid & \cdots & \mathbf{a}_{n} \end{array}\right]\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array}\right]=x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+\cdots+x_{n} \mathbf{a}_{n} .

c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0}

3(1,-1,0)+2(-2,1,2)+(1,1,-4)=(0,0,0) .

\begin{aligned} & (1,-1,0)=-\frac{2}{3}(-2,1,2)-\frac{1}{3}(1,1,-4), \\ & (-2,1,2)=-\frac{3}{2}(1,-1,0)-\frac{1}{2}(1,1,-4), \quad \text { and } \\ & (1,1,-4)=-3(1,-1,0)-2(-2,1,2) . \end{aligned}

2(1,2,1)-(2,4,2)+0(2,1,1)=(0,0,0) .

\left[\begin{array}{ccc} 1 & -2 & 1 \\ -1 & 1 & 1 \\ 0 & 2 & -4 \end{array}\right]

\left[\begin{array}{ccc} 1 & 1 & 0 \\ 2 & 0 & -1 \\ 3 & 1 & 2 \end{array}\right]

\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{array}\right]

V=\left[\begin{array}{ccccc} 1 & a_{0} & a_{0}^{2} & \cdots & a_{0}^{n} \\ 1 & a_{1} & a_{1}^{2} & \cdots & a_{1}^{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & a_{n} & a_{n}^{2} & \cdots & a_{n}^{n} \end{array}\right],

c_{0}\left[\begin{array}{c} 1 \\ 1 \\ \vdots \\ 1 \end{array}\right]+c_{1}\left[\begin{array}{c} a_{0} \\ a_{1} \\ \vdots \\ a_{n} \end{array}\right]+c_{2}\left[\begin{array}{c} a_{0}^{2} \\ a_{1}^{2} \\ \vdots \\ a_{n}^{2} \end{array}\right]+\cdots+c_{n}\left[\begin{array}{c} a_{0}^{n} \\ a_{1}^{n} \\ \vdots \\ a_{n}^{n} \end{array}\right]=\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 0 \end{array}\right]

p(x)=c_{n} x^{n}+\cdots+c_{2} x^{2}+c_{1} x+c_{0}

\left[\begin{array}{ccccc} 1 & x_{0} & x_{0}^{2} & \cdots & x_{0}^{n} \\ 1 & x_{1} & x_{1}^{2} & \cdots & x_{1}^{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n} & x_{n}^{2} & \cdots & x_{n}^{n} \end{array}\right]\left[\begin{array}{c} c_{0} \\ c_{1} \\ \vdots \\ c_{n} \end{array}\right]=\left[\begin{array}{c} y_{0} \\ y_{1} \\ \vdots \\ y_{n} \end{array}\right],

\left[\begin{array}{ccc} 1 & -2 & 4 \\ 1 & 1 & 1 \\ 1 & 3 & 9 \end{array}\right]\left[\begin{array}{l} c_{0} \\ c_{1} \\ c_{2} \end{array}\right]=\left[\begin{array}{c} 13 \\ -2 \\ 8 \end{array}\right],

p(x)=c_{2} x^{2}+c_{1} x+c_{0}=2 x^{2}-3 x-1 .

\{(1,2,3),(-1, k, 1),(1,1,0)\}

A_{2}=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad \text { and } \quad A_{3}=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right] .

c_{1} \mathbf{v}_{1}+\cdots+c_{k} \mathbf{v}_{k} \in \mathcal{S}

\left[\begin{array}{ccccc} 3 & -1 & 1 & 0 & -1 \\ 2 & 1 & -1 & 4 & 0 \\ 1 & 1 & -1 & 1 & 2 \\ 2 & 1 & 2 & 0 & 0 \\ -1 & 3 & -1 & 3 & 2 \end{array}\right]

\left[\begin{array}{ccccccc} 4 & 1 & 2 & 3 & 2 & 2 & -1 \\ 0 & -1 & 4 & 2 & 0 & 4 & 2 \\ -1 & 2 & 0 & 2 & 4 & 3 & 3 \\ 3 & 1 & -1 & 0 & 2 & 2 & 3 \\ -1 & 0 & -1 & 4 & 3 & 0 & 3 \end{array}\right]

\operatorname{span}(\mathbf{v}, \mathbf{w})=\operatorname{span}(\mathbf{v}, \mathbf{v}+\mathbf{w}) .

\operatorname{span}\left(\mathbf{w}, \mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right)=\operatorname{span}\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right) .

V_{n}=\left[\begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 4 & \cdots & 2^{n-1} \\ 1 & 3 & 9 & \cdots & 3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & n & n^{2} & \cdots & n^{n-1} \end{array}\right] .

c_{1} \mathbf{e}_{1}+c_{2} \mathbf{e}_{2}+\cdots+c_{n} \mathbf{e}_{n}=\mathbf{0} .

\begin{aligned} {\left[\begin{array}{lll|l} 1 & 1 & 2 & x \\ 1 & 2 & 1 & y \\ 2 & 1 & 1 & z \end{array}\right] } & \xrightarrow{\substack{R_{2}-R_{1} \\ R_{3}-2 R_{1}}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & x \\ 0 & 1 & -1 & y-x \\ 0 & -1 & -3 & z-2 x \end{array}\right] \\ & \xrightarrow{R_{3}+R_{2}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & x \\ 0 & 1 & -1 & y-x \\ 0 & 0 & -4 & z+y-3 x \end{array}\right] . \end{aligned}

\begin{aligned} {\left[\begin{array}{lll|l} 1 & 1 & 2 & 0 \\ 1 & 2 & 1 & 0 \\ 2 & 1 & 1 & 0 \end{array}\right] } & \xrightarrow{R_{2}-R_{1}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & 0 \\ R_{3}-2 R_{1} & 1 & -1 & 0 \\ 0 & -1 & -3 & 0 \end{array}\right] \\ & \xrightarrow{R_{3}+R_{2}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & -4 & 0 \end{array}\right] . \end{aligned}

|B| \leq|C| .

\mathbf{v}_{i}=a_{i, 1} \mathbf{w}_{1}+a_{i, 2} \mathbf{w}_{2}+\cdots+a_{i, \ell} \mathbf{w}_{\ell} \quad \text { for } \quad 1 \leq i \leq k,

a_{i, 1} \mathbf{w}_{1}+a_{i, 2} \mathbf{w}_{2}+\cdots+a_{i, \ell} \mathbf{w}_{\ell},

V \mathbf{x}=\left(W A^{T}\right) \mathbf{x}=W\left(A^{T} \mathbf{x}\right)=W \mathbf{0}=\mathbf{0} .

\begin{aligned} & {\left[\begin{array}{lll|l} 1 & 1 & 3 & 0 \\ 1 & 2 & 2 & 0 \\ 1 & 3 & 1 & 0 \end{array}\right] \xrightarrow{\substack{R_{2}-R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{ccc|c} 1 & 1 & 3 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 2 & -2 & 0 \end{array}\right] } \\ & \xrightarrow{R_{3}-2 R_{2}}\left[\begin{array}{ccc|c} 1 & 1 & 3 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

\begin{aligned} {\left[\begin{array}{cc|c} 1 & 3 & x \\ -1 & 3 & y \\ -1 & -1 & z \end{array}\right] } & \xrightarrow{\substack{R_{2}+R_{1} \\ R_{3}+R_{1}}}\left[\begin{array}{cc|c} 1 & 3 & x \\ 0 & 6 & x+y \\ 0 & 2 & x+z \end{array}\right] \\ & \xrightarrow{R_{3}-\frac{1}{3} R_{2}}\left[\begin{array}{cc|c} 1 & 3 & x \\ 0 & 6 & x+y \\ 0 & 0 & \frac{2}{3} x-\frac{1}{3} y+z \end{array}\right] . \end{aligned}

d \mathbf{w}+c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0},

c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0} .

\begin{aligned} & {\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ -1 & 2 & 1 & 0 \\ -1 & -1 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_{2}+R_{1} \\ R_{3}+R_{1}}}\left[\begin{array}{lll|l} 1 & 2 & 1 & 0 \\ 0 & 4 & 2 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right] } \\ & \xrightarrow{R_{3}-\frac{1}{4} R_{2}}\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 4 & 2 & 0 \\ 0 & 0 & 1 / 2 & 0 \end{array}\right] . \end{aligned}

3(1)-(-1)+4(-1)=0 \quad \text { and } \quad 3(2)-(2)+4(-1)=0 .

d \mathbf{w}+c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{m} \mathbf{v}_{m}=\mathbf{0} .

\mathbf{w}=-\frac{c_{1}}{d} \mathbf{v}_{1}-\frac{c_{2}}{d} \mathbf{v}_{2}-\cdots-\frac{c_{m}}{d} \mathbf{v}_{m} .

\operatorname{range}(A)=\operatorname{span}((1,2,1),(1,-1,0),(2,1,1)) .

\begin{aligned} & {\left[\begin{array}{ccc|c} 1 & 1 & 2 & 0 \\ 2 & -1 & 1 & 0 \\ 1 & 0 & 1 & 0 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & 0 \\ 0 & -3 & -3 & 0 \\ 0 & -1 & -1 & 0 \end{array}\right] } \\ & \xrightarrow{\frac{-1}{3} R_{2}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & -1 & 0 \end{array}\right] \xrightarrow{\substack{R_{1}-R_{2} \\ R_{3}+R_{2}}}\left[\begin{array}{lll|l} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

A=\left[\begin{array}{ccc} 1 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 0 & 1 \end{array}\right] \quad \text { is } \quad\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right],

A=\left[\begin{array}{ccccc} 1 & 0 & 1 & 0 & -1 \\ 1 & 1 & 0 & 0 & 1 \\ -1 & 0 & -1 & 1 & 4 \\ 2 & 1 & 1 & -1 & -3 \end{array}\right]

\begin{aligned} & {\left[\begin{array}{ccccc} 1 & 0 & 1 & 0 & -1 \\ 1 & 1 & 0 & 0 & 1 \\ -1 & 0 & -1 & 1 & 4 \\ 2 & 1 & 1 & -1 & -3 \end{array}\right] \xrightarrow{\substack{R_{2}-R_{1} \\ R_{3}+R_{1} \\ R_{4}-2 R_{1}}}\left[\begin{array}{ccccc} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & -1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 1 & -1 & -1 & -1 \end{array}\right]} \\ & \xrightarrow{R_{4}-R_{2}}\left[\begin{array}{ccccc} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & -1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & -1 & -3 \end{array}\right] \xrightarrow{R_{4}+R_{3}}\left[\begin{array}{ccccc} 1 & 0 & 1 & 0 & -1 \\ 0 & 1 & -1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=x_{3}(-1,1,1,0,0)+x_{5}(1,-2,0,-3,1) .

\left[\begin{array}{lllllll} 1 & 2 & 0 & 3 & 4 & 0 & 5 \\ 0 & 0 & 1 & 6 & 7 & 0 & 8 \\ 0 & 0 & 0 & 0 & 0 & 1 & 9 \end{array}\right]

\left[\begin{array}{l} 2 \\ 0 \\ 0 \end{array}\right] \rightarrow\left[\begin{array}{c} 2 \\ -1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right], \quad\left[\begin{array}{l} 3 \\ 6 \\ 0 \end{array}\right] \rightarrow\left[\begin{array}{c} 3 \\ 0 \\ 6 \\ -1 \\ 0 \\ 0 \\ 0 \end{array}\right], \quad\left[\begin{array}{l} 4 \\ 7 \\ 0 \end{array}\right] \rightarrow\left[\begin{array}{c} 4 \\ 0 \\ 7 \\ 0 \\ -1 \\ 0 \\ 0 \end{array}\right], \quad\left[\begin{array}{l} 5 \\ 8 \\ 9 \end{array}\right] \rightarrow\left[\begin{array}{c} 5 \\ 0 \\ 8 \\ 0 \\ 0 \\ 9 \\ -1 \end{array}\right]

\begin{aligned} & \{(2,-1,0,0,0,0,0),(3,0,6,-1,0,0,0) \\ & \qquad(4,0,7,0,-1,0,0),(5,0,8,0,0,9,-1)\} \end{aligned}

\operatorname{range}(A), \operatorname{null}(A), \operatorname{range}\left(A^{T}\right), \text { and } \operatorname{null}\left(A^{T}\right)

A=\left[\begin{array}{ccccc} 1 & 0 & 1 & 0 & -1 \\ 1 & 1 & 0 & 0 & 1 \\ -1 & 0 & -1 & 1 & 4 \\ 2 & 1 & 1 & -1 & -3 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{cccc} 1 & 1 & -1 & 2 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & -1 & 1 \\ 0 & 0 & 1 & -1 \\ -1 & 1 & 4 & -3 \end{array}\right] } & \xrightarrow{\substack{R_{3}-R_{1} \\ R_{5}+R_{1}}}\left[\begin{array}{cccc} 1 & 1 & -1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 2 & 3 & -1 \end{array}\right] \\ \xrightarrow{\substack{R_{1}-R_{2} \\ R_{3}+R_{2} \\ R_{5}-2 R_{2}}}\left[\begin{array}{cccc} 1 & 0 & -1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 3 & -3 \end{array}\right] & \xrightarrow{R_{3} \leftrightarrow R_{4}}\left[\begin{array}{cccc} 1 & 0 & -1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & -3 \end{array}\right] \\ & \xrightarrow{\substack{R_{1}+R_{3} \\ R_{5}-3 R_{3}}}\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

\{(1,0,1,0,-1),(1,1,0,0,1),(-1,0,-1,1,4)\} .

\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=x_{4}(0,-1,1,1),

\operatorname{range}\left(A^{T}\right)=\operatorname{span}\left(\mathbf{a}_{1}, \mathbf{a}_{2}, \ldots, \mathbf{a}_{m}\right) .

A^{T} E^{T}=\left[A^{T} \mathbf{v}_{1}\left|A^{T} \mathbf{v}_{2}\right| \cdots \mid A^{T} \mathbf{v}_{m}\right]=R^{T} .

\begin{figure} \[ \begin{aligned} {[A \mid I] } & =\left[\begin{array}{cccccccc|cccc} * & * & * & * & * & * & * & * & 1 & 0 & 0 & 0 \\ * & * & * & * & * & * & * & * & 0 & 1 & 0 & 0 \\ * & * & * & * & * & * & * & * & 0 & 0 & 1 & 0 \\ * & * & * & * & * & * & * & * & 0 & 0 & 0 & 1 \end{array}\right] \\ \xrightarrow{\text { row-reduce }} & (t)\left[\begin{array}{llllllll|llll} \star & * & * & * & * & * & * & * & * & * & * & * \\ 0 & 0 & \star & * & * & * & * & * & * & * & * & * \\ 0 & 0 & 0 & 0 & 0 & \star & * & * & * & * & * & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & * & * & * & * \end{array}\right]=[R \mid E] \leftarrow \operatorname{null}\left(A^{T}\right) \\ & \operatorname{range}\left(A^{T}\right) \end{aligned} \] \captionsetup{labelformat=empty} \caption{Figure 2.22: After we row-reduce \([A \mid I]\) to a row echelon form \([R \mid E]\), we can immediately read off bases of \(\operatorname{range}\left(A^{T}\right)\) and \(\operatorname{null}\left(A^{T}\right)\) from the rows of that matrix.} \end{figure}

\begin{gathered} {\left[\begin{array}{cccc|ccc} 1 & 1 & 1 & -1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ -1 & 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{3}+R_{1}}\left[\begin{array}{cccc|ccc} 1 & 1 & 1 & -1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 2 & 0 & 1 & 0 & 1 \end{array}\right]} \\ \xrightarrow{R_{1}-R_{2}} \\ {\left[\begin{array}{cccc|ccc} 1 & 0 & 0 & -1 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \end{array}\right] .} \end{gathered}

\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=x_{3}(0,-1,1,0)+x_{4}(1,0,0,1) .

A=\left[\begin{array}{ccc} 1 & 1 & 2 \\ 2 & -1 & 1 \\ 1 & 0 & 1 \end{array}\right]

\begin{array}{r} {\left[\begin{array}{ccccc} 0 & 0 & -2 & 2 & -2 \\ 2 & -2 & -1 & 3 & 3 \\ -1 & 1 & -1 & 0 & -3 \end{array}\right] \xrightarrow{R_{1} \leftrightarrow R_{3}}\left[\begin{array}{ccccc} -1 & 1 & -1 & 0 & -3 \\ 2 & -2 & -1 & 3 & 3 \\ 0 & 0 & -2 & 2 & -2 \end{array}\right]} \\ \xrightarrow{R_{2}+2 R_{1}}\left[\begin{array}{ccccc} -1 & 1 & -1 & 0 & -3 \\ 0 & 0 & -3 & 3 & -3 \\ 0 & 0 & -2 & 2 & -2 \end{array}\right] \xrightarrow{R_{3}-\frac{2}{3} R_{2}}\left[\begin{array}{ccccc} -1 & 1 & -1 & 0 & -3 \\ 0 & 0 & -3 & 3 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{array}

\operatorname{rank}(A)+\operatorname{nullity}(A)=r+(n-r)=n,

A=\left[\begin{array}{ccccc} 1 & 0 & -1 & 0 & 0 \\ 2 & 1 & -1 & 2 & 2 \\ -1 & -1 & 0 & -2 & -2 \\ 1 & 0 & -1 & 0 & 1 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{ccccc} 1 & 0 & -1 & 0 & 0 \\ 2 & 1 & -1 & 2 & 2 \\ 1 & 0 & -1 & 0 & 1 \\ -1 & -1 & 0 & -2 & -2 \end{array}\right] } & \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-R_{1} \\ R_{4}+R_{1}}}\left[\begin{array}{ccccc} 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & -1 & -1 & -2 & -2 \end{array}\right] \\ & \xrightarrow{R_{4}+R_{2}}\left[\begin{array}{ccccc} 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

\operatorname{span}\left(\mathbf{a}_{1}+\mathbf{b}_{1}, \mathbf{a}_{2}+\mathbf{b}_{2}, \ldots, \mathbf{a}_{n}+\mathbf{b}_{n}\right) \subseteq \operatorname{span}\left(\mathbf{a}_{1}, \mathbf{b}_{1}, \mathbf{a}_{2}, \mathbf{b}_{2}, \ldots, \mathbf{a}_{n}, \mathbf{b}_{n}\right)

A B=\left[A \mathbf{b}_{1}\left|A \mathbf{b}_{2}\right| \cdots \mid A \mathbf{b}_{p}\right]=\left[\sum_{j=1}^{n} b_{j, 1} \mathbf{a}_{j}\left|\sum_{j=1}^{n} b_{j, 2} \mathbf{a}_{j}\right| \cdots \mid \sum_{j=1}^{n} b_{j, p} \mathbf{a}_{j}\right] .

\left[\begin{array}{cccccccc} 0 & 0 & 2 & 2 & -1 & -2 & 3 & -1 \\ -6 & -12 & -1 & -7 & 1 & 8 & -7 & 6 \\ 4 & 8 & -2 & 2 & 1 & -2 & 1 & -2 \\ -1 & -2 & 0 & -1 & 0 & 1 & -1 & 1 \end{array}\right]

\left[\begin{array}{ccccccc} 5 & 1 & 9 & 6 & -1 & 1 & 8 \\ 2 & 2 & 2 & 4 & 1 & 0 & 0 \\ 5 & 4 & 6 & 9 & 1 & 1 & 4 \\ 6 & 3 & 9 & 9 & 0 & 1 & 7 \\ -3 & 1 & -7 & -2 & 2 & -1 & -8 \end{array}\right]

A_{2}=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad \text { and } \quad A_{3}=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right] .

A=\left[\begin{array}{ccccc} x & 1 & 1 & \cdots & 1 \\ 1 & x & 1 & \cdots & 1 \\ 1 & 1 & x & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & x \end{array}\right] \in \mathcal{M}_{n} .

B=\{(1,-1,2),(-1,2,3),(2,-2,4)\}

\begin{array}{llll} 0+0=0 & 0+1=1 & 1+0=1 & 1+1=0, \quad \text { and } \\ 0 \times 0=0 & 0 \times 1=0 & 1 \times 0=0 & 1 \times 1=1 . \end{array}

\begin{aligned} x_{1}+x_{3}+x_{5} & =1 \\ x_{1}+x_{2}+x_{5} & =0 \\ x_{3}+x_{4} & =1 \\ x_{1}+x_{3}+x_{4}+x_{5} & =0 \\ x_{2}+x_{3}+x_{5} & =1 \end{aligned}

\begin{aligned} & {\left[\begin{array}{lllll|l} 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 1 \end{array}\right] \xrightarrow{R_{2}-R_{1}}\left[\begin{array}{lllll|l} 1 & 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 & 1 \end{array}\right] } \\ & \xrightarrow{R_{5}-R_{2}} {\left[\begin{array}{lllll|l} 1 & 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right] \xrightarrow{R_{1}-R_{5}}\left[\begin{array}{llll|l} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] } \\ & \xrightarrow{R_{3}-R_{4}} {\left[\begin{array}{lllll|l} 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right] \xrightarrow{R_{1}-R_{3}}\left[\begin{array}{llll|l} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] . } \end{aligned}

\begin{aligned} x_{1}+x_{3}+x_{4} & =1 \\ x_{2}+x_{3} & =1 \\ x_{1}+x_{3}+x_{4}+x_{5} & =0 \\ x_{1}+x_{2}+x_{5} & =0 \\ x_{2}+x_{3}+x_{5} & =0 \end{aligned}

\begin{aligned} {\left[\begin{array}{lllll|l} 1 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \end{array}\right] \xrightarrow{\substack{R_{3}-R_{1} \\ R_{4}-R_{1}}}\left[\begin{array}{lllll|l} 1 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \end{array}\right] } \\ \xrightarrow{\substack{R_{4}-R_{2} \\ R_{5}-R_{2}}}\left[\begin{array}{lllll|l} 1 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \end{array}\right] \xrightarrow{R_{3} \leftrightarrow R_{4}}\left[\begin{array}{llll|l} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \\ \xrightarrow{\substack{R_{3}-R_{4} \\ R_{5}-R_{4}}}\left[\begin{array}{lllll|l} 1 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{1}-R_{3}}\left[\begin{array}{llll|l} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=(0,1,0,1,1) \quad \text { and } \quad\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=(1,0,1,1,1) .

\begin{aligned} \quad(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) & \quad(\text { all buttons start "off") } \\ +(1,1,1,0,0,1,0,0,0,0,0,0,0,0,0,0) & \quad(\text { press } \# 1 \text { toggles } 1,2,3,6) \\ +(0,0,1,0,0,1,1,1,0,0,1,0,0,0,0,0) & \quad(\text { press } \# 2 \text { toggles } 3,6,7,8,11) \\ =(1,1,0,0,0,0,1,1,0,0,1,0,0,0,0,0) & \quad(\text { just add, but } 1+1=0) \end{aligned}

\begin{aligned} & \mathbf{v}_{\mathrm{s}}=(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) \quad \text { and } \\ & \mathbf{v}_{\mathrm{e}}=(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) . \end{aligned}

\begin{aligned} \mathbf{v}_{\mathrm{s}}+\left(x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+\cdots+x_{16} \mathbf{a}_{16}\right) & =\mathbf{v}_{\mathrm{e}}, \quad \text { or } \\ x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+\cdots+x_{16} \mathbf{a}_{16} & =\mathbf{v}_{\mathrm{e}}-\mathbf{v}_{\mathrm{s}} . \end{aligned}

A=\left[\begin{array}{cccc|cccc|cccc|cccc} 1 & 1 & . & . & 1 & . & . & . & . & . & . & . & . & . & . & . \\ 1 & 1 & 1 & . & . & 1 & . & . & . & . & . & . & . & . & . & . \\ . & 1 & 1 & 1 & . & . & 1 & . & . & . & . & . & . & . & . & . \\ . & . & 1 & 1 & . & . & . & 1 & . & . & . & . & . & . & . & . \\ \hline 1 & . & . & . & 1 & 1 & . & . & 1 & . & . & . & . & . & . & . \\ . & 1 & . & . & 1 & 1 & 1 & . & . & 1 & . & . & . & . & . & . \\ . & . & 1 & . & . & 1 & 1 & 1 & . & . & 1 & . & . & . & . & . \\ . & . & . & 1 & . & . & 1 & 1 & . & . & . & 1 & . & . & . & . \\ \hline . & . & . & . & 1 & . & . & . & 1 & 1 & . & . & 1 & . & . & . \\ . & . & . & . & . & 1 & . & . & 1 & 1 & 1 & . & . & 1 & . & . \\ . & . & . & . & . & . & 1 & . & . & 1 & 1 & 1 & . & . & 1 & . \\ . & . & . & . & . & . & . & 1 & . & . & 1 & 1 & . & . & . & 1 \\ \hline . & . & . & . & . & . & . & . & 1 & . & . & . & 1 & 1 & . & . \\ . & . & . & . & . & . & . & . & . & 1 & . & . & 1 & 1 & 1 & . \\ . & . & . & . & . & . & . & . & . & . & 1 & . & . & 1 & 1 & 1 \\ . & . & . & . & . & . & . & . & . & . & . & 1 & . & . & 1 & 1 \end{array}\right] .

\left[\begin{array}{cccc|cccc|cccc|cccc|c} 1 & . & . & . & . & . & . & . & . & . & . & . & . & 1 & 1 & 1 & 1 \\ . & 1 & . & . & . & . & . & . & . & . & . & . & 1 & 1 & . & 1 & 1 \\ . & . & 1 & . & . & . & . & . & . & . & . & . & 1 & . & 1 & 1 & 1 \\ . & . & . & 1 & . & . & . & . & . & . & . & . & 1 & 1 & 1 & . & 1 \\ \hline . & . & . & . & 1 & . & . & . & . & . & . & . & 1 & . & 1 & . & 1 \\ . & . & . & . & . & 1 & . & . & . & . & . & . & . & . & . & 1 & . \\ . & . & . & . & . & . & 1 & . & . & . & . & . & 1 & . & . & . & . \\ . & . & . & . & . & . & . & 1 & . & . & . & . & . & 1 & . & 1 & 1 \\ \hline . & . & . & . & . & . & . & . & 1 & . & . & . & 1 & 1 & . & . & 1 \\ . & . & . & . & . & . & . & . & . & 1 & . & . & 1 & 1 & 1 & . & 1 \\ . & . & . & . & . & . & . & . & . & . & 1 & . & . & 1 & 1 & 1 & 1 \\ . & . & . & . & . & . & . & . & . & . & . & 1 & . & . & 1 & 1 & 1 \\ \hline . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \end{array}\right],

\mathbf{x}=(0,0,1,0,1,0,0,0,0,0,0,1,0,1,0,0) .

\mathbf{y}=\left[\begin{array}{l} 0 \\ 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \quad \text { and } \quad A=\left[\begin{array}{lllll} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \end{array}\right] \quad \text { then } \quad B=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{array}\right] .

R=\left[\begin{array}{llllll} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \quad \text { then } \quad \operatorname{null}(A)=\operatorname{span}\left(\left[\begin{array}{l} 0 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \end{array}\right]\right) .

\begin{aligned} w+y+2 z & =2 \\ x+y+z & =1 \\ 2 w+x+y+z & =0 \\ w+y & =1 \end{aligned}

\begin{aligned} & {\left[\begin{array}{llll|l} 1 & 0 & 1 & 2 & 2 \\ 0 & 1 & 1 & 1 & 1 \\ 2 & 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 \end{array}\right] \xrightarrow{\substack{R_{3}-2 R_{1} \\ R_{4}-R_{1}}}\left[\begin{array}{llll|l} 1 & 0 & 1 & 2 & 2 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 0 & 2 \\ 0 & 0 & 0 & 1 & 2 \end{array}\right] } \\ & \xrightarrow{R_{3}-R_{2}} {\left[\begin{array}{llll|l} 1 & 0 & 1 & 2 & 2 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 \end{array}\right] \xrightarrow{\substack{R_{1}-2 R_{4} \\ R_{2}-R_{4} \\ R_{3}-2 R_{4}}}\left[\begin{array}{llll|l} 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 \end{array}\right] } \\ & \xrightarrow{\substack{R_{1}-R_{3} \\ R_{2}-R_{3}}}\left[\begin{array}{llll|l} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 \end{array}\right] . \end{aligned}

0+0+0=0, \quad 1+1+1=0, \quad 2+2+2=0, \quad \text { and } \quad 0+1+2=0 .

\begin{aligned} 2 w+4 y & =1 \\ 3 w+x+2 y & =0 \\ 4 x+4 y+z & =1 \\ 3 w+3 x+4 y+2 z & =1 \end{aligned}

\begin{aligned} {\left[\begin{array}{llll|l} 2 & 0 & 4 & 0 & 1 \\ 3 & 1 & 2 & 0 & 0 \\ 0 & 4 & 4 & 1 & 1 \\ 3 & 3 & 4 & 2 & 1 \end{array}\right] \xrightarrow{3 R_{1}}\left[\begin{array}{llll|l} 1 & 0 & 2 & 0 & 3 \\ 3 & 1 & 2 & 0 & 0 \\ 0 & 4 & 4 & 1 & 1 \\ 3 & 3 & 4 & 2 & 1 \end{array}\right] } \\ \xrightarrow{\substack{R_{2}-3 R_{1} \\ R_{4}-3 R_{1}}}\left[\begin{array}{llll|l} 1 & 0 & 2 & 0 & 3 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 4 & 4 & 1 & 1 \\ 0 & 3 & 3 & 2 & 2 \end{array}\right] \xrightarrow{\substack{R_{3}-4 R_{2} \\ R_{4}-3 R_{2}}}\left[\begin{array}{llll|l} 1 & 0 & 2 & 0 & 3 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 2 & 4 \end{array}\right] \\ \xrightarrow{R_{4}-2 R_{3}}\left[\begin{array}{llll|l} 1 & 0 & 2 & 0 & 3 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

(3,1,0,2),(1,0,1,2),(4,4,2,2),(2,3,3,2), \text { and }(0,2,4,2) .

\begin{aligned} & x+y+z=1 \\ & x+y=1 \\ & x+z=0 \end{aligned}

\begin{aligned} & x+y+z=0 \\ & x+y=1 \end{aligned}

\begin{array}{r} w+x+z=1 \\ w+y+z=0 \\ x+y+z=1 \end{array}

\begin{array}{r} x+z=0 \\ w+y+z=1 \\ w+x+y+z=0 \\ x+y+z=1 \end{array}

\begin{aligned} v+w+x & =0 \\ w+x+y & =1 \\ x+y+z & =0 \\ v+y+z & =1 \\ v+w+z & =0 \end{aligned}

\begin{aligned} r+s+w+x & =1 \\ r+v+x+y & =0 \\ s+v+w+z & =1 \\ r+w+y+z & =1 \\ s+w+x+z & =0 \end{aligned}

\begin{array}{r} x+2 y+2 z=0 \\ 2 x+y+2 z=1 \\ x+z=2 \end{array}

\begin{aligned} 2 w+x+y & =2 \\ w \quad+2 y+2 z & =1 \\ x+y+2 z & =0 \\ 2 x+2 y+z & =0 \end{aligned}

\begin{array}{r} x+2 y+3 z=4 \\ 2 x+3 y+z=1 \\ 4 x+y+2 z=0 \end{array}

\begin{array}{r} 4 w+y+3 z=3 \\ 2 w+x+y+2 z=1 \\ w+3 x+4 z=3 \\ 4 x+y+3 z=3 \end{array}

\begin{aligned} & v 3 w+x+y+z \text { is even } \\ & -2 v-w+2 x+y \text { is odd } \\ & v+w \quad+y+z \text { is odd } \\ & w-x+2 y-z \text { is even, and } \\ & v+w-3 x+y+2 z \text { is even. } \end{aligned}

\begin{array}{rrrrrrrr} v & + & w & + & x & + & y & - \\ v & - & 2 w & + & 2 x & + & 2 y & + \\ -v & + & w & + & x & & & + \\ & & & + & z & \\ 2 w & & - & x & + & y & - & 3 z \\ 2 v & & & - & 2 y & + & z & - \\ & & & & \end{array}

\begin{aligned} \operatorname{maximize}: & x_{1}+3 x_{2} \\ \text { subject to: } & 2 x_{1}+x_{2} \leq 4 \\ & x_{1}+3 x_{2} \leq 6 \\ & x_{1}, \quad x_{2} \geq 0 \end{aligned}

\begin{array}{ll} \operatorname{maximize}: & \mathbf{c} \cdot \mathbf{x} \\ \text { subject to: } & A \mathbf{x} \leq \mathbf{b} \\ & \mathbf{x} \geq \mathbf{0} \end{array}

A=\left[\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right], \quad \mathbf{b}=(4,6), \quad \text { and } \quad \mathbf{c}=(1,3),

\begin{aligned} \operatorname{maximize}: & \mathbf{c} \cdot \mathbf{x} \\ \text { subject to: } & A \mathbf{x} \leq \mathbf{b} \\ & \mathbf{x} \geq \mathbf{0} \end{aligned}

\begin{array}{lr} \operatorname{minimize}: & -x_{1}-2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{cc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

A=\left[\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right], \quad \mathbf{b}=(3,1), \quad \text { and } \quad \mathbf{c}=(1,2) .

\begin{array}{lc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & 3 x_{1}-4 x_{2} \geq 1 \\ & 2 x_{1}+x_{2}=5 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{aligned} & \text { maximize: } x_{1}+2 x_{2} \\ & \text { subject to: }-3 x_{1}+4 x_{2} \leq-1 \\ & 2 x_{1}+x_{2} \leq 5 \\ &-2 x_{1}-x_{2} \leq-5 \\ & x_{1}, \quad x_{2} \geq 0 \end{aligned}

A=\left[\begin{array}{cc} -3 & 4 \\ 2 & 1 \\ -2 & -1 \end{array}\right], \quad \mathbf{b}=(-1,5,-5), \quad \text { and } \quad \mathbf{c}=(1,2) .

\begin{array}{lc} \operatorname{maximize}: & 2 x_{1}-x_{2} \\ \text { subject to: } & x_{1}+3 x_{2} \leq 4 \\ & 3 x_{1}-2 x_{2} \leq 5 \\ & x_{1} \geq 0 \end{array}

\begin{array}{lc} \operatorname{maximize}: & 2 x_{1}-\left(x_{2}^{+}-x_{2}^{-}\right) \\ \text {subject to: } & x_{1}+3\left(x_{2}^{+}-x_{2}^{-}\right) \leq 4 \\ & 3 x_{1}-2\left(x_{2}^{+}-x_{2}^{-}\right) \leq 5 \\ & x_{1}, \quad x_{2}^{+}, \quad x_{2}^{-} \geq 0 \end{array}

A=\left[\begin{array}{ccc} 1 & 3 & -3 \\ 3 & -2 & 2 \end{array}\right], \quad \mathbf{b}=(4,5), \quad \text { and } \quad \mathbf{c}=(2,-1,1) .

\begin{array}{rlrl} \operatorname{minimize}: & 3 x_{1}-2 x_{2} & \\ \text { subject to: } & x_{1}+3 x_{2} & =4 \\ x_{1}+2 x_{2} & \geq 3 \\ x_{2} & \geq 0 \end{array}

\begin{array}{cc} \operatorname{maximize}: & -3 x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+3 x_{2} \leq 4 \\ - & x_{1}-3 x_{2} \leq-4 \\ - & x_{1}-2 x_{2} \leq-3 \\ & x_{2} \geq 0 \end{array}

\begin{array}{lr} \operatorname{maximize}: & -3\left(x_{1}^{+}-x_{1}^{-}\right)+2 x_{2} \\ \text { subject to: } & x_{1}^{+}-x_{1}^{-}+3 x_{2} \leq 4 \\ & -\left(x_{1}^{+}-x_{1}^{-}\right)-3 x_{2} \leq-4 \\ & -\left(x_{1}^{+}-x_{1}^{-}\right)-2 x_{2} \leq-3 \\ & x_{1}^{+}, \quad x_{1}^{-}, \quad x_{2} \geq 0 \end{array}

A=\left[\begin{array}{ccc} 1 & -1 & 3 \\ -1 & 1 & -3 \\ -1 & 1 & -2 \end{array}\right], \quad \mathbf{b}=(4,-4,-3), \quad \text { and } \quad \mathbf{c}=(-3,3,2) .

\begin{array}{lc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{lr} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{cc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}-x_{2} \leq-1 \\ & -x_{1}+x_{2} \leq-1 \\ & x_{1}, x_{2} \geq 0 \end{array}

\begin{array}{ll} \operatorname{maximize}: & 2 x_{1}+x_{2} \\ \text { subject to: } & 3 x_{1}+2 x_{2} \geq-1 \\ & x_{1}-x_{2} \leq 2 \\ & x_{1}-3 x_{2} \geq-3 \\ & x_{1} \geq 0 \end{array}

\begin{array}{lrl} \text { maximize: } & 2 x_{1}+x_{2}-x_{3} & \\ \text { subject to: } & -3 x_{1}-2 x_{2}+2 x_{3} & \leq 1 \\ & x_{1}-x_{2}+x_{3} & \leq 2 \\ - & x_{1}+3 x_{2}-3 x_{3} & \leq 3 \\ & x_{1}, \quad x_{2}, \quad x_{3} & \geq 0 \end{array}

\begin{array}{lrlrl} \operatorname{maximize}: & & & 2 x_{1}+x_{2}-x_{3} & \\ \text { subject to: } & s_{1} & & -3 x_{1}-2 x_{2}+2 x_{3} & =1 \\ & & s_{2} & +x_{1}-x_{2}+x_{3} & =2 \\ & & & s_{3}-x_{1}+3 x_{2}-3 x_{3} & =3 \\ & s_{1}, & s_{2}, & s_{3}, \quad x_{1}, \quad x_{2}, \quad x_{3} & \geq 0 \end{array}

\begin{array}{llllr} \operatorname{maximize}: & z & & & \\ \text { subject to: } & z & & & -2 x_{1}-x_{2}+x_{3}=0 \\ & & s_{1} & & -3 x_{1}-2 x_{2}+2 x_{3}=1 \\ & & & s_{2} & +x_{1}-x_{2}+x_{3}=2 \\ & & & s_{3}-x_{1}+3 x_{2}-3 x_{3}=3 \\ & & s_{1}, & s_{2}, & s_{3}, \quad x_{1}, \quad x_{2}, \quad x_{3} \geq \end{array}

\begin{aligned} \operatorname{maximize}: & \mathbf{c} \cdot \mathbf{x} \\ \text { subject to: } & A \mathbf{x} \leq \mathbf{b} \\ & \mathbf{x} \geq \mathbf{0} \end{aligned}

\left[\begin{array}{c|c|c|c} 1 & \mathbf{0}^{T} & -\mathbf{c}^{T} & 0 \\ \hline \mathbf{0} & I & A & \mathbf{b} \end{array}\right] .

z-\mathbf{c} \cdot \mathbf{x}=0\left[\begin{array}{c|ccc|ccc|c} z & s_{1} & s_{2} & s_{3} & x_{1} & x_{2} & x_{3} & \\ \hline 1 & 0 & 0 & 0 & -2 & -1 & 1 & 0 \\ \hline 0 & 1 & 0 & 0 & -3 & -2 & 2 & 1 \\ 0 & 0 & 1 & 0 & 1 & -1 & 1 & 2 \\ 0 & 0 & 0 & 1 & -1 & 3 & -3 & 3 \end{array}\right],

x_{1}=x_{2}=x_{3}=0, \quad s_{1}=1, s_{2}=2, s_{3}=3, \quad z=0 .

\begin{aligned} -3 x_{1}-2 x_{2}+2 x_{3} & =-3 x_{1} \leq 1 \\ x_{1}-x_{2}+x_{3} & =x_{1} \leq 2 \\ -x_{1}+3 x_{2}-3 x_{3} & =-x_{1} \leq 3 . \end{aligned}

\left[\begin{array}{c|ccc|ccc|c} 1 & 0 & 0 & 0 & -2 & -1 & 1 & 0 \\ \hline 0 & 1 & 0 & 0 & -3 & -2 & 2 & 1 \\ 0 & 0 & 1 & 0 & 1 & -1 & 1 & 2 \\ 0 & 0 & 0 & 1 & -1 & 3 & -3 & 3 \end{array}\right]

\xrightarrow{\substack{R_{1}+2 R_{3} \\ R_{2}+3 R_{3} \\ R_{4}+R_{3}}}\left[\begin{array}{c|ccc|ccc|c} 1 & 0 & 2 & 0 & 0 & -3 & 3 & 4 \\ \hline 0 & 1 & 3 & 0 & 0 & -5 & 5 & 7 \\ 0 & 0 & 1 & 0 & 1 & -1 & 1 & 2 \\ 0 & 0 & 1 & 1 & 0 \uparrow & 2 & -2 & 5 \end{array}\right] .

s_{2}=x_{2}=x_{3}=0, \quad s_{1}=7, x_{1}=2, s_{3}=5, \quad z=4 .

\begin{gathered} {\left[\begin{array}{c|ccc|ccc|c} 1 & 0 & 2 & 0 & 0 & -3 & 3 & 4 \\ \hline 0 & 1 & 3 & 0 & 0 & -3 & 5 & 7 \\ 0 & 0 & 1 & 0 & 1 & -1 & 1 & 2 \\ 0 & 0 & 1 & 1 & 0 & 2 & -2 & 5 \end{array}\right]} \\ 5 / 2 \text { is smallest (and only) positive ratio } \end{gathered}

\begin{aligned} & \begin{array}{lllllll} z & s_{1} & s_{2} & s_{3} & x_{1} & x_{2} & x_{3} \end{array} \\ & {\left[\begin{array}{c|ccc|ccc|c} 1 & 0 & 2 & 0 & 0 & -3 & 3 & 4 \\ \hline 0 & 1 & 3 & 0 & 0 & -5 & 5 & 7 \\ 0 & 0 & 1 & 0 & 1 & -1 & 1 & 2 \\ 0 & 0 & 1 & 1 & 0 & 2 & -2 & 5 \end{array}\right] \begin{array}{c} \text { new "z" value } \\ \downarrow \end{array}} \\ & \xrightarrow{\substack{(1 / 2) R_{4} \\ R_{1}+3 R_{4} \\ R_{2}+5 R_{4} \\ R_{3}+R_{4}}}\left[\begin{array}{c|ccc|ccc|c} 1 & 0 & 7 / 2 & 3 / 2 & 0 & 0 & 0 & 23 / 2 \\ \hline 0 & 1 & 11 / 2 & 5 / 2 & 0 & 0 & 0 & 39 / 2 \\ 0 & 0 & 3 / 2 & 1 / 2 & 1 & 0 & 0 & 9 / 2 \\ 0 & 0 & 1 / 2 & 1 / 2 & 0 & 1 & -1 & 5 / 2 \end{array}\right] \\ & \text { ↑ new "leading" entry } \end{aligned}

x_{\mathrm{p}}+0.8 x_{\mathrm{c}}+1.5 x_{\mathrm{pc}},

x_{\mathrm{c}}+x_{\mathrm{pc}} \leq 100, \quad x_{\mathrm{p}}+x_{\mathrm{pc}} \leq 80, \quad \text { and } \quad x_{\mathrm{p}}+x_{\mathrm{c}}+x_{\mathrm{pc}} \leq 150,

\begin{array}{rlrl} \operatorname{maximize}: & & x_{\mathrm{p}}+0.8 x_{\mathrm{c}}+1.5 x_{\mathrm{pc}} & \\ \text { subject to: } & & x_{\mathrm{c}}+ & x_{\mathrm{pc}} \\ & x_{\mathrm{p}} & +100 \\ & x_{\mathrm{p}}+ & x_{\mathrm{c}}+ & x_{\mathrm{pc}} \\ & x_{\mathrm{p}}, & x_{\mathrm{c}}, & x_{\mathrm{pc}} \geq 0 \end{array}

\begin{array}{lr} \operatorname{maximize}: & x_{1}+3 x_{2} \\ \text { subject to: } & x_{1}+2 x_{2} \leq 4 \\ & -2 x_{1}+x_{2} \leq 0 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{cc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 2 \\ & -x_{1}-x_{2} \leq-1 \\ & -2 x_{1}+x_{2} \leq-2 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{aligned} \operatorname{maximize}: & -y \\ \text { subject to: } & x_{1}+x_{2} \\ & -x_{1}-x_{2}-y \leq 2 \\ & -2 x_{1}+x_{2}-y \leq-2 \\ & x_{1}, x_{2}, y \geq 0 \end{aligned}

\begin{array}{r} \\ \left.\xrightarrow{\begin{array}{c} z \\ R_{1} \end{array}} \begin{array}{c} S_{2} \quad S_{3} \\ R_{3}+R_{4} \\ R_{4}+R_{4} \\ \hline 0 \\ 0 \\ 0 \end{array} \left\lvert\, \begin{array}{c|cccc|c} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \hline \end{array}\left[\begin{array}{c|ccc|ccc} 1 & 0 & 0 & 1 & -2 & 1 & 0 \\ \hline 0 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & -1 & 1 & -2 & 0 \\ 0 & 0 & 0 & -1 & 2 & -1 & 1 \end{array}\right] \begin{array}{c} 2 \\ \text { pivot } \end{array}\right.\right] \end{array}

\left[\begin{array}{c|ccc|cc|c} 1 & 1 & 0 & 0 & 0 & -1 & 2 \\ \hline 0 & 1 & 0 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 & 3 & 2 \end{array}\right]

\begin{array}{lc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\left(x_{1}+x_{2}\right)+\left(-x_{1}+x_{2}\right)=2 x_{2} \leq 3+1=4, \quad \text { so } \quad x_{2} \leq 2 .

x_{2}+\left(x_{1}+x_{2}\right)=x_{1}+2 x_{2} \leq 2+3=5 .

y_{1}\left(x_{1}+x_{2}\right)+y_{2}\left(-x_{1}+x_{2}\right) \leq 3 y_{1}+y_{2} .

x_{1}+2 x_{2} \leq y_{1}\left(x_{1}+x_{2}\right)+y_{2}\left(-x_{1}+x_{2}\right),

\begin{aligned} \operatorname{maximize}: & \mathbf{c} \cdot \mathbf{x} \\ \text { subject to: } & A \mathbf{x} \leq \mathbf{b} \\ & \mathbf{x} \geq \mathbf{0} \end{aligned}

\begin{array}{rr} \operatorname{minimize}: & \mathbf{b} \cdot \mathbf{y} \\ \text { subject to: } & A^{T} \mathbf{y} \geq \mathbf{c} \\ & \mathbf{y} \geq \mathbf{0} \end{array}

\mathbf{c} \cdot \mathbf{x} \leq \mathbf{b} \cdot \mathbf{y} .

\begin{array}{cc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

A=\left[\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right], \quad \mathbf{b}=(3,1), \quad \text { and } \quad \mathbf{c}=(1,2),

\begin{array}{cc} \operatorname{minimize}: & 3 y_{1}+y_{2} \\ \text { subject to: } & y_{1}-y_{2} \geq 1 \\ & y_{1}+y_{2} \geq 2 \\ & y_{1}, \quad y_{2} \geq 0 \end{array}

\begin{array}{lcl} \text { maximize: } & x_{1}+x_{2}+x_{3}+x_{4}+x_{5} & \\ \text { subject to: } & x_{1}+x_{2}+x_{3} & \leq 3 \\ & x_{2}+x_{3}+x_{4} & \leq 3 \\ & x_{3}+x_{4}+x_{5} & \leq 3 \\ & x_{1}+x_{2}+x_{5} & \leq 3 \\ & x_{1}, x_{2}, x_{3}, x_{4}, x_{5} & \geq 0 \end{array}

x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=1+1+1+1+1=5

\begin{array}{rlrl} \operatorname{minimize}: & 3 y_{1}+3 y_{2}+3 y_{3}+3 y_{4}+3 y_{5} & \\ \text { subject to: } & y_{1}+y_{4}+y_{5} & \geq 1 \\ & y_{1}+y_{2}+y_{5} & \geq 1 \\ & y_{1}+y_{2}+y_{3} & \geq 1 \\ & y_{2}+y_{3}+y_{4} & \geq 1 \\ & y_{3}+y_{4}+y_{5} \geq 1 \\ y_{2}, \quad y_{3}, \quad y_{4}, \quad y_{5} & \geq 0 \end{array}

\mathbf{c} \cdot \mathbf{x}_{*}=\mathbf{b} \cdot \mathbf{y}_{*} .

\begin{array}{lc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2}=3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{cc} \operatorname{maximize}: & x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 3 \\ & -x_{1}-x_{2} \leq-3 \\ & -x_{1}+x_{2} \leq 1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{ll} \operatorname{minimize}: & 3 y_{1}-3 y_{2}+y_{3} \\ \text { subject to: } & y_{1}-y_{2}-y_{3} \geq 1 \\ & y_{1}-y_{2}+y_{3} \geq 2 \\ & y_{1}, \quad y_{2}, \quad y_{3} \geq 0 \end{array}

\begin{array}{lr} \operatorname{minimize}: & 3 y_{*}+y_{3} \\ \text { subject to: } & y_{*}-y_{3} \geq 1 \\ & y_{*}+y_{3} \geq 2 \\ & y_{3} \geq 0 \end{array}

\begin{aligned} & \geq \\ & \leq \\ & \leq \end{aligned}

\begin{aligned} & \leq 0 \\ & \text { unc } \\ & \geq 0 \end{aligned}

unconstrained

\begin{aligned} & \geq 0 \\ & \text { unconstrained } \\ & \leq 0 \end{aligned}

\begin{aligned} & \geq \\ & = \\ & \leq \end{aligned}

\begin{array}{lc} \operatorname{maximize}: & x_{2} \\ \text { subject to: } & x_{1}+x_{2} \\ & 2 x_{1}-x_{2} \geq 1 \\ & x_{1} \geq 0 \end{array}

\begin{array}{lr} \operatorname{minimize}: & -3 y_{1}+y_{2} \\ \text { subject to: } & y_{1}+2 y_{2} \geq 0 \\ & y_{1}-y_{2}=1 \\ & y_{2} \leq 0 \end{array}

\begin{aligned} \operatorname{maximize}: & x_{1}+x_{2} \\ \text { subject to: } & 2 x_{1}+x_{2} \geq 1 \\ & x_{1}+2 x_{2} \leq 3 \\ & x_{1}, \quad x_{2} \geq 0 \end{aligned}

\begin{array}{lr} \operatorname{minimize}: & 2 x_{1}-x_{2} \\ \text { subject to: } & x_{1}+x_{2} \leq 4 \\ 2 x_{1}+3 x_{2} \geq 4 & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{aligned} \operatorname{minimize}: \quad 3 x_{1}+x_{2} \\ \text { subject to: } \quad \begin{aligned} x_{1}+x_{2} & \geq 2 \\ 2 x_{1}-x_{2} & \geq 1 \\ x_{1}, x_{2} & \geq 0 \end{aligned} \end{aligned}

\begin{aligned} \operatorname{maximize}: \quad 2 x_{1}+x_{2} & \\ \text { subject to: } \quad x_{1}+x_{2} & \geq 2 \\ x_{1}+2 x_{2} & \geq 3 \\ 2 x_{1}+2 x_{2} & \leq 5 \\ x_{1}, \quad x_{2} & \geq 0 \end{aligned}

\begin{array}{lr} \operatorname{maximize}: & 4 x_{1}+x_{2} \\ \text { subject to: } & 2 x_{1}-2 x_{2} \leq 5 \\ & x_{1}+3 x_{2} \leq 3 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{lr} \operatorname{maximize}: & x_{1}+x_{2}+x_{3} \\ \text { subject to: } & x_{1}+2 x_{2}+3 x_{3} \leq 2 \\ & 3 x_{1}+2 x_{2}+x_{3} \leq 1 \\ & x_{1}, \quad x_{2}, \quad x_{3} \geq 0 \end{array}

\begin{array}{lrl} \operatorname{maximize}: & x_{1}+x_{2}+x_{3} & \\ \text { subject to: } & 2 x_{1}-x_{2}+x_{3} & \leq 1 \\ & x_{1}+3 x_{2} & \leq 2 \\ - & x_{1}+x_{2}+2 x_{3} & \leq 3 \\ & x_{1}, \quad x_{2}, \quad x_{3} & \geq 0 \end{array}

\begin{array}{lc} \operatorname{maximize}: & 2 x_{1}+x_{2}+x_{3}+x_{4} \\ \text { subject to: } & x_{1}+2 x_{2}-3 x_{3}+2 x_{4} \leq 1 \\ & 2 x_{1}-x_{2}+x_{3}+x_{4} \leq 0 \\ & 3 x_{1}+x_{2}-2 x_{3}-x_{4} \leq 1 \\ & x_{1}, \quad x_{2}, \quad x_{3}, \quad x_{4} \geq 0 \end{array}

\begin{aligned} \operatorname{maximize}: & x_{1}+x_{2}+x_{3}+x_{4} \\ \text { subject to: } & 4 x_{1}+3 x_{2}+2 x_{3} \\ x_{1}+4 x_{2}+x_{3}+2 x_{4} & \leq 2 \\ 2 x_{1}+x_{2}+3 x_{3}+2 x_{4} & \leq 2 \\ x_{1}, \quad x_{2}, \quad x_{3}, \quad x_{4} & \geq 0 \end{aligned}

\begin{array}{lr} \operatorname{maximize}: & x_{2} \\ \text { subject to: } & 2 x_{1}-2 x_{2} \geq 2 \\ x_{1}+3 x_{2} \leq 3 \\ x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{lr} \operatorname{minimize}: & 2 x_{1}-x_{2}-2 x_{3} \\ \text { subject to: } & 2 x_{1}-x_{2}+4 x_{3} \geq 2 \\ & x_{1}+2 x_{2}-3 x_{3} \leq 1 \\ & x_{1}, \quad x_{2}, \quad x_{3} \geq 0 \end{array}

\begin{aligned} \operatorname{minimize}: & x_{1}+2 x_{2}-x_{3} \\ \text { subject to: } & x_{2}-x_{3} \leq 3 \\ 2 x_{1}+2 x_{2}-x_{3} & =1 \\ x_{1}+2 x_{2}+2 x_{3} & \leq 3 \\ x_{1}, \quad x_{2}, \quad x_{3} & \geq 0 \end{aligned}

\begin{aligned} \operatorname{maximize}: \quad 3 x_{1}+2 x_{2}-x_{3}+2 x_{4} \\ \text { subject to: } \quad 2 x_{1}-2 x_{2}-x_{3}+2 x_{4} \geq 2 \\ x_{1}+3 x_{2}-2 x_{3}+x_{4}=2 \\ 3 x_{1}+x_{2}-3 x_{3}+3 x_{4} \leq 3 \\ x_{1}, \quad x_{2}, \quad x_{4} \geq 0 \end{aligned}

\begin{aligned} \operatorname{maximize}: & x_{1}+2 x_{2}+x_{3}+2 x_{4} \\ \text { subject to: } & 2 x_{1}+4 x_{2}-x_{3} \\ 3 x_{2}+x_{3}+2 x_{4} & =1 \\ 3 x_{1}-x_{2}+2 x_{3}+2 x_{4} & \leq 3 \\ x_{1}, \quad x_{3} & \geq 0 \end{aligned}

\begin{array}{lc} \operatorname{maximize}: & x_{1}+x_{2} \\ \text { subject to: } & 2 x_{1}+x_{2} \leq 2 \\ & 2 x_{1}+3 x_{2} \geq 3 \\ & x_{1}-2 x_{2} \geq-1 \\ & x_{1}, \quad x_{2} \geq 0 \end{array}

\begin{array}{lr} \operatorname{maximize}: & x_{2} \\ \text { subject to: } & c x_{1}+x_{2} \leq c \\ & -c x_{1}+x_{2} \leq 0 \\ & x_{1}, x_{2} \geq 0 \end{array}

\|\mathbf{v}\|_{1} \stackrel{\text { def }}{=} \sum_{j=1}^{n}\left|v_{j}\right| \quad \text { and } \quad\|\mathbf{v}\|_{\infty} \stackrel{\text { def }}{=} \max _{1 \leq j \leq n}\left\{\left|v_{j}\right|\right\},

\operatorname{rank}(A B) \leq \min \{\operatorname{rank}(A), \operatorname{rank}(B)\} .

\operatorname{rank}(A)=\operatorname{rank}(C R) \leq \min \{\operatorname{rank}(C), \operatorname{rank}(R)\} \leq \min \{m, r, n\} \leq r .

\begin{figure} \[ \left.A=\left[\begin{array}{llllll} * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \end{array}\right]=\left[\begin{array}{lll} * & * & * \\ * & * & * \\ * & * & * \\ * & * & * \\ * & * & * \\ * & * & * \\ * & * & * \end{array}\right]\left[\begin{array}{cccccc} * & * & * & * & * & * \\ * & * & * & * & * & * \\ * & * & * & * & * & * \end{array}\right]\right\} \] \captionsetup{labelformat=empty} \caption{Figure 2.39: Theorem 2.C. 1 says that every matrix can be written as a product of a tall and skinny matrix and a short and fat matrix. The rank determines exactly how skinny and short the matrices in the product can be (as shown here, \(A\) has rank 3, so it can be written as a product of a matrix \(C\) with 3 columns and a matrix \(R\) with 3 rows).} \end{figure}

P=[C \mid D] \quad \text { and } \quad \hat{R}=\left[\frac{R}{O}\right],

A=P \hat{R}=[C \mid D]\left[\frac{R}{O}\right]=C R+D O=C R,

\left[\begin{array}{cccc|ccc} 1 & 0 & 0 & -1 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \end{array}\right] .

\hat{R}=\left[\begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right], \quad E=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{array}\right], \quad \text { and } \quad P=\left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ -1 & 1 & 1 \end{array}\right] .

A=C R, \quad \text { where } \quad C=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{array}\right] \quad \text { and } \quad R=\left[\begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \end{array}\right] .

A=\left[\begin{array}{ccccccccccc} 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 & 3 & 0 & 1 & 0 & 2 & 0 \\ 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & 4 & 0 & 2 & 0 & 6 & 0 & 2 & 0 & 4 & 0 \\ 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & -2 & 0 & -1 & 0 & -3 & 0 & -1 & 0 & -2 & 0 \\ 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & 4 & 0 & 2 & 0 & 6 & 0 & 2 & 0 & 4 & 0 \\ 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 & 3 & 0 & 1 & 0 & 2 & 0 \\ 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \end{array}\right]

\begin{aligned} C^{T} & =\left[\begin{array}{ccccccccccc} 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 2 & 0 & -1 & 0 & 2 & 0 & 1 & 0 \end{array}\right] \quad \text { and } \\ R & =\left[\begin{array}{ccccccccccc} 1 & 0 & -1 & 0 & 2 & 0 & 2 & 0 & -1 & 0 & 1 \\ 0 & 2 & 0 & 1 & 0 & 3 & 0 & 1 & 0 & 2 & 0 \end{array}\right], \end{aligned}

A=\sum_{j=1}^{r} \mathbf{v}_{j} \mathbf{w}_{j}^{T}

C=\left[\mathbf{v}_{1}\left|\mathbf{v}_{2}\right| \cdots \mid \mathbf{v}_{r}\right] \quad \text { and } \quad R^{T}=\left[\mathbf{w}_{1}\left|\mathbf{w}_{2}\right| \cdots \mid \mathbf{w}_{r}\right] .

A=C R=\left[\mathbf{v}_{1}\left|\mathbf{v}_{2}\right| \cdots \mid \mathbf{v}_{r}\right]\left[\begin{array}{c} \frac{\mathbf{w}_{1}^{T}}{\mathbf{w}_{2}^{T}} \\ \hline \vdots \\ \hline \mathbf{w}_{n}^{T} \end{array}\right]=\sum_{j=1}^{r} \mathbf{v}_{j} \mathbf{w}_{j}^{T},

\operatorname{rank}(A)=\operatorname{rank}\left(\sum_{j=1}^{r} \mathbf{v}_{j} \mathbf{w}_{j}^{T}\right) \leq \sum_{j=1}^{r} \operatorname{rank}\left(\mathbf{v}_{j} \mathbf{w}_{j}^{T}\right)=\sum_{j=1}^{r} 1=r,

A=\left[\begin{array}{cccc} 1 & 1 & 1 & -1 \\ 0 & 1 & 1 & 0 \\ -1 & 1 & 1 & 1 \end{array}\right] .

A=C R, \quad \text { where } \quad C=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{array}\right] \quad \text { and } \quad R=\left[\begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \end{array}\right] .

A=\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]\left[\begin{array}{llll} 1 & 0 & 0 & -1 \end{array}\right]+\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]\left[\begin{array}{llll} 0 & 1 & 1 & 0 \end{array}\right] .

A=\left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \end{array}\right]

A=\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{array}\right]

\operatorname{rank}(A)=\operatorname{dim}(\operatorname{range}(A))=\operatorname{dim}\left(\operatorname{span}\left(\mathbf{a}_{1}, \mathbf{a}_{2}, \ldots, \mathbf{a}_{n}\right)\right) .

\operatorname{rank}(C)=\operatorname{dim}\left(\operatorname{range}\left(C^{T}\right)\right)=\operatorname{dim}\left(\operatorname{span}\left(\mathbf{c}_{1}, \mathbf{c}_{2}, \ldots, \mathbf{c}_{m}\right)\right) .

\left[\begin{array}{llll} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right] \quad \rightarrow \quad\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] .

\left[\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 & 16 \\ 1 & 2 & 5 & 1 & 4 \\ 1 & 3 & 9 & 27 & 81 \end{array}\right] \quad \longrightarrow \quad\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{array}\right] .

\left[\begin{array}{cccccc} 4 & 5 & 0 & 4 & 1 & 3 \\ 2 & 2 & 1 & 2 & 1 & 1 \\ -4 & 0 & 3 & -2 & -3 & 1 \\ 3 & 5 & 4 & 4 & 1 & 3 \end{array}\right]

\left[\begin{array}{ccccccc} -1 & 0 & 3 & -1 & 2 & -1 & -2 \\ 3 & 1 & -3 & 2 & -2 & 1 & 4 \\ 3 & 2 & 3 & 1 & 2 & -1 & 2 \\ 6 & 6 & 2 & 0 & 2 & 0 & 6 \\ 6 & 4 & 6 & 2 & 4 & -2 & 4 \end{array}\right]

D=\left[\begin{array}{cc} I_{\operatorname{rank}(A)} & O \\ O & O \end{array}\right] .

V=\left[\begin{array}{ccccc} 1 & a_{0} & a_{0}^{2} & \cdots & a_{0}^{n} \\ 1 & a_{1} & a_{1}^{2} & \cdots & a_{1}^{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & a_{m} & a_{m}^{2} & \cdots & a_{m}^{n} \end{array}\right] .

A=\left[\begin{array}{ccc} 1 & -2 & -1 \\ 2 & -1 & -1 \\ 3 & 6 & 2 \end{array}\right]

\begin{aligned} {\left[\begin{array}{ccc|ccc} 1 & -2 & -1 & 1 & 0 & 0 \\ 2 & -1 & -1 & 0 & 1 & 0 \\ 3 & 6 & 2 & 0 & 0 & 1 \end{array}\right] } & \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-3 R_{1}}}\left[\begin{array}{ccc|ccc} 1 & -2 & -1 & 1 & 0 & 0 \\ 0 & 3 & 1 & -2 & 1 & 0 \\ 0 & 12 & 5 & -3 & 0 & 1 \end{array}\right] \\ & \xrightarrow{R_{3}-4 R_{2}}\left[\begin{array}{ccc|ccc} 1 & -2 & -1 & 1 & 0 & 0 \\ 0 & 3 & 1 & -2 & 1 & 0 \\ 0 & 0 & 1 & 5 & -4 & 1 \end{array}\right] . \end{aligned}

R=\left[\begin{array}{ccc} 1 & -2 & -1 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{array}\right] \quad \text { and } \quad P=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 5 & -4 & 1 \end{array}\right]^{-1}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{array}\right] .

\left[\begin{array}{llll} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \end{array}\right] .

\begin{aligned} & {\left[\begin{array}{cccc|ccc} 2 & 1 & -2 & 1 & 1 & 0 & 0 \\ -4 & -4 & 3 & 0 & 0 & 1 & 0 \\ 2 & -5 & -2 & 8 & 0 & 0 & 1 \end{array}\right] } \\ & \xrightarrow{\substack{R_{2}+2 R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{cccc|ccc} 2 & 1 & -2 & 1 & 1 & 0 & 0 \\ 0 & -2 & -1 & 2 & 2 & 1 & 0 \\ 0 & -6 & 0 & 7 & -1 & 0 & 1 \end{array}\right] \\ & \xrightarrow{R_{3}-3 R_{2}}\left[\begin{array}{cccc|ccc} 2 & 1 & -2 & 1 & 1 & 0 & 0 \\ 0 & -2 & -1 & 3 & 2 & 1 & 0 \\ 0 & 0 & 3 & 1 & -7 & -3 & 1 \end{array}\right] \end{aligned}

U=\left[\begin{array}{cccc} 2 & 1 & -2 & 1 \\ 0 & -2 & -1 & 3 \\ 0 & 0 & 3 & 1 \end{array}\right] \text { and } L=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 0 \\ -7 & -3 & 1 \end{array}\right]^{-1}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & 3 & 1 \end{array}\right] .

A=\left[\begin{array}{ll} 2 & 4 \\ 2 & 7 \end{array}\right] \quad \text { then } \quad A=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 2 & 4 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 2 & 3 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]

\mathbf{a}_{i}^{T}=\mathbf{u}_{i}^{T}+\sum_{j=1}^{i-1} \ell_{i, j} \mathbf{u}_{j}^{T} \quad \text { for all } \quad 1 \leq i \leq m .

R_{m}+\ell_{m, 1} R_{1}, \quad R_{m}+\ell_{m, 2} R_{2}, \quad \ldots, \quad R_{m}+\ell_{m, m-1} R_{m-1}

\mathbf{u}_{m}^{T}+\ell_{m, 1} \mathbf{u}_{1}^{T}+\ell_{m, 2} \mathbf{u}_{2}^{T}+\cdots+\ell_{m, m-1} \mathbf{u}_{m-1}^{T},

\begin{aligned} & R_{m}+\ell_{m, m-1} R_{m-1}, \\ & R_{m}+\ell_{m, m-2} R_{m-2}, \\ & \quad \vdots \\ & R_{m}+\ell_{m, 1} R_{1}, \end{aligned} \quad R_{m-1}+\ell_{m-1, m-2} R_{m-2}=\frac{}{} .

A=\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 4 & 9 & 0 & -1 \\ -6 & -9 & 7 & 6 \\ -2 & -2 & 9 & 0 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 4 & 9 & 0 & -1 \\ -6 & -9 & 7 & 6 \\ -2 & -2 & 9 & 0 \end{array}\right] } & \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}+3 R_{1} \\ R_{4}+R_{1}}}\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 0 & 1 & 2 & 1 \\ 0 & 3 & 4 & 3 \\ 0 & 2 & 8 & -1 \end{array}\right] \\ & \xrightarrow{R_{3}-3 R_{2}}\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 4 & -3 \end{array}\right] \xrightarrow{R_{4}+2 R_{3}}\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -3 \end{array}\right] . \end{aligned}

L=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ -3 & 3 & 1 & 0 \\ -1 & 2 & -2 & 1 \end{array}\right] \quad \text { and } \quad U=\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -3 \end{array}\right] .

A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right],

A \mathbf{x}=L U \mathbf{x}=L(U \mathbf{x})=L \mathbf{y}=\mathbf{b},

\left[\begin{array}{cccc} 2 & 4 & -1 & -1 \\ 4 & 9 & 0 & -1 \\ -6 & -9 & 7 & 6 \\ -2 & -2 & 9 & 0 \end{array}\right]\left[\begin{array}{c} w \\ x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \\ 1 \end{array}\right]

\begin{aligned} & {\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 \\ -3 & 3 & 1 & 0 & 0 \\ -1 & 2 & -2 & 1 & 1 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}+3 R_{1} \\ R_{4}+R_{1}}}\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 3 & 1 & 0 & 0 \\ 0 & 2 & -2 & 1 & 1 \end{array}\right]} \\ & \xrightarrow{\substack{R_{3}-3 R_{2} \\ R_{4}-2 R_{2}}}\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & -6 \\ 0 & 0 & -2 & 1 & -3 \end{array}\right] \xrightarrow{R_{4}+2 R_{3}}\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & -6 \\ 0 & 0 & 0 & 1 & -15 \end{array}\right], \end{aligned}

\begin{gathered} {\left[\begin{array}{cccc|c} 2 & 4 & -1 & -1 & 0 \\ 0 & 1 & 2 & 1 & 2 \\ 0 & 0 & -2 & 0 & -6 \\ 0 & 0 & 0 & -3 & -15 \end{array}\right] \xrightarrow{\substack{R_{1} / 2 \\ -R_{3} / 2 \\ -R_{4} / 3}}\left[\begin{array}{cccc|c} 1 & 2 & -1 / 2 & -1 / 2 & 0 \\ 0 & 1 & 2 & 1 & 2 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 5 \end{array}\right]} \\ \xrightarrow{R_{1}+\frac{1}{2} R_{4}}\left[\begin{array}{cccc|c} 1 & 2 & -1 / 2 & 0 & 5 / 2 \\ 0 & 1 & 2 & 0 & -3 \\ R_{2}-R_{4} & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 5 \end{array}\right] \xrightarrow{R_{1}+\frac{1}{2} R_{3}}\left[\begin{array}{cccc|c} 1 & 2 & 0 & 0 & 4 \\ 0 & 1 & 0 & 0 & -9 \\ R_{2}-2 R_{3} & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 5 \end{array}\right] \\ \xrightarrow{R_{1}-2 R_{2}}\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 22 \\ 0 & 1 & 0 & 0 & -9 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 5 \end{array}\right] . \end{gathered}

\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \end{array}\right]\left[\begin{array}{l} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{array}\right]=\left[\begin{array}{c} 3 \\ -1 \\ -3 \\ 9 \end{array}\right] .

\begin{aligned} {\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \end{array}\right] } & \xrightarrow{\substack{R_{2}-R_{1} \\ R_{3}-R_{1} \\ R_{4}-R_{1}}}\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 2 & 4 & 8 \\ 0 & 3 & 9 & 27 \end{array}\right] \\ & \xrightarrow{\substack{R_{3}-2 R_{2} \\ R_{4}-3 R_{2}}}\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 6 \\ 0 & 0 & 6 & 24 \end{array}\right] \xrightarrow{R_{4}-3 R_{3}}\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & 6 \end{array}\right] . \end{aligned}

L=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \end{array}\right] \quad \text { and } \quad U=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & 6 \end{array}\right] .

p(x)=2 x^{3}-5 x^{2}-x+3 .

\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \end{array}\right]\left[\begin{array}{l} c_{0} \\ c_{1} \\ c_{2} \\ c_{3} \end{array}\right]=\left[\begin{array}{c} -1 \\ -1 \\ -1 \\ 5 \end{array}\right] .

p(x)=x^{3}-3 x^{2}+2 x-1 .

A \mathbf{x}=\mathbf{b}, \quad A \mathbf{y}=\mathbf{c}, \quad \text { and } \quad A \mathbf{z}=\mathbf{d},

\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]

A=\left[\begin{array}{llll} 0 & 2 & 4 & 2 \\ 2 & 1 & 1 & 2 \\ 1 & 1 & 2 & 3 \\ 1 & 2 & 3 & 4 \end{array}\right]

\begin{gathered} {\left[\begin{array}{llll|llll} 0 & 2 & 4 & 2 & 1 & 0 & 0 & 0 \\ 2 & 1 & 1 & 2 & 0 & 1 & 0 & 0 \\ 1 & 1 & 2 & 3 & 0 & 0 & 1 & 0 \\ 1 & 2 & 3 & 4 & 0 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_{1} \leftrightarrow R_{3}}\left[\begin{array}{llll|llll} 1 & 1 & 2 & 3 & 0 & 0 & 1 & 0 \\ 2 & 1 & 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 2 & 4 & 2 & 1 & 0 & 0 & 0 \\ 1 & 2 & 3 & 4 & 0 & 0 & 0 & 1 \end{array}\right]} \\ \xrightarrow{R_{2}-2 R_{1}}\left[\begin{array}{cccc|cccc} 1 & 1 & 4 & 3 & 0 & 0 & 1 & 0 \\ 0 & -1 & -3 & -4 & 0 & 1 & -2 & 0 \\ 0 & 2 & 4 & 2 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & -1 & 1 \end{array}\right] \\ \xrightarrow{R_{4}-R_{1}+2 R_{2}}\left[\begin{array}{cccc|cccc} 1 & 1 & 4 & 3 & 0 & 0 & 1 & 0 \\ 0 & -1 & -3 & -4 & 0 & 1 & -2 & 0 \\ 0 & 0 & -2 & -6 & 1 & 2 & -4 & 0 \\ 0 & 0 & -2 & -3 & 0 & 1 & -3 & 1 \end{array}\right] \\ \xrightarrow{R_{4}-R_{3}}\left[\begin{array}{cccc|cccc} 1 & 1 & 4 & 3 & 0 & 0 & 1 & 0 \\ 0 & -1 & -3 & -4 & 0 & 1 & -2 & 0 \\ 0 & 0 & -2 & -6 & 1 & 2 & -4 & 0 \\ 0 & 0 & 0 & 3 & -1 & -1 & 1 & 1 \end{array}\right] . \end{gathered}

\begin{aligned} R & =\left[\begin{array}{cccc} 1 & 1 & 4 & 3 \\ 0 & -1 & -3 & -4 \\ 0 & 0 & -2 & -6 \\ 0 & 0 & 0 & 3 \end{array}\right] \text { and } \\ E^{-1} & =\left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 1 & 2 & -4 & 0 \\ -1 & -1 & 1 & 1 \end{array}\right]^{-1}=\left[\begin{array}{cccc} 0 & -2 & 1 & 0 \\ 2 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & -1 & 1 & 1 \end{array}\right] . \end{aligned}

\left[\begin{array}{cccc} 0 & -2 & 1 & 0 \\ 2 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & -1 & 1 & 1 \end{array}\right] \xrightarrow{R_{1} \leftrightarrow R_{3}}\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 0 & -2 & 1 & 0 \\ 1 & -1 & 1 & 1 \end{array}\right],

E^{-1}=P L, \quad \text { where } \quad P=\left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \text { and } \quad L=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 0 & -2 & 1 & 0 \\ 1 & -1 & 1 & 1 \end{array}\right] .

A=P L U

E=\left[\begin{array}{ccccc} 0 & 0 & \star & 0 & 0 \\ 0 & 0 & * & \star & 0 \\ \star & 0 & * & * & 0 \\ * & 0 & * & * & \star \\ * & \star & * & * & * \end{array}\right] .

E^{-1}=P_{*}^{-1} L_{*}^{-1} .

E=\left[\begin{array}{ccccc} 0 & 0 & \star & 0 & 0 \\ 0 & 0 & * & \star & 0 \\ \star & 0 & * & * & 0 \\ * & 0 & * & * & \star \\ * & \star & * & * & * \end{array}\right] \text { then } P^{T}=\left[\begin{array}{ccccc} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \end{array}\right] .

[U \mid E]=\left[\begin{array}{cccc|cccc} 1 & 1 & 2 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & -1 & 0 & 1 & 0 \\ 0 & 0 & 2 & 1 & 1 & 0 & -1 & 1 \\ 0 & 0 & 0 & 3 / 2 & -1 / 2 & 1 & -1 / 2 & 1 / 2 \end{array}\right] .

P^{T}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array}\right] \quad \text { and } \quad E^{-1}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 0 & -1 / 2 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right] .

\begin{aligned} L=P^{-1} E^{-1}=P^{T} E^{-1} & =\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array}\right]\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 0 & -1 / 2 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right] \\ & =\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & -1 / 2 & 1 \end{array}\right] . \end{aligned}

A \mathbf{x}=P L U \mathbf{x}=P L(U \mathbf{x})=P L \mathbf{y}=P(L \mathbf{y})=P \mathbf{c}=\mathbf{b},

\mathbf{c}=P^{T} \mathbf{b}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} 0 \\ 1 \\ 2 \\ 3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 3 \\ 1 \end{array}\right] .

\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 2 \\ 0 & 1 & 1 & 0 & 3 \\ 1 & 0 & -1 / 2 & 1 & 1 \end{array}\right] \xrightarrow{\text { row-reduce }}\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 3 / 2 \end{array}\right],

\left[\begin{array}{cccc|c} 1 & 1 & 2 & 1 & 0 \\ 0 & 1 & -1 & 0 & 2 \\ 0 & 0 & 2 & 1 & 1 \\ 0 & 0 & 0 & 3 / 2 & 3 / 2 \end{array}\right] \xrightarrow{\text { row-reduce }}\left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & -3 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \end{array}\right],

A=\left[\begin{array}{c|c} B & \mathbf{v} \\ \hline \mathbf{w}^{T} & c \end{array}\right] .

A=\left[\begin{array}{c|c} B & \mathbf{v} \\ \hline \mathbf{w}^{T} & c \end{array}\right]=\underbrace{\left[\begin{array}{c|c} L & \mathbf{0} \\ \hline \mathbf{x}^{T} & 1 \end{array}\right]}_{\text {"new" } L} \underbrace{\left[\begin{array}{c|c} U & \mathbf{y} \\ \hline \mathbf{0}^{T} & d \end{array}\right]}_{\text {"new" } U}=\left[\begin{array}{c|c} L U & L \mathbf{y} \\ \hline \mathbf{x}^{T} U & d+\mathbf{x}^{T} \mathbf{y} \end{array}\right] .

\operatorname{rank}\left(\left[\frac{B}{\mathbf{w}^{T}}\right]\right)>r,

[3], \quad\left[\begin{array}{ll} 3 & -2 \\ 3 & -1 \end{array}\right], \quad \text { and } \quad\left[\begin{array}{ccc} 3 & -2 & -1 \\ 3 & -1 & 4 \\ 1 & 5 & 2 \end{array}\right] .

A=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]

A=\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] .

A=\left[\begin{array}{cccccc} 2 & -1 & 0 & 0 & 0 & 0 \\ -1 & 3 & -1 & 0 & 1 & 0 \\ 0 & -1 & 4 & -1 & 0 & 0 \\ 0 & 0 & -1 & 4 & -1 & 0 \\ 0 & 0 & 0 & -1 & 3 & -1 \\ 0 & 1 & 0 & 0 & -1 & 2 \end{array}\right] .

\left[\begin{array}{ll} 2 & -1 \\ 4 & -1 \end{array}\right] .

V=\left[\begin{array}{ccccc} 1 & a_{0} & a_{0}^{2} & \cdots & a_{0}^{n} \\ 1 & a_{1} & a_{1}^{2} & \cdots & a_{1}^{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & a_{n} & a_{n}^{2} & \cdots & a_{n}^{n} \end{array}\right],

A=\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right],

\begin{aligned} (2,1) & =2(1,0)+(0,1) \\ & =-(0,1)+2(1,1) \\ & =(1,0)+(1,1), \end{aligned}

\begin{gathered} \mathbf{v}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k} \quad \text { and } \\ \mathbf{v}=d_{1} \mathbf{v}_{1}+d_{2} \mathbf{v}_{2}+\cdots+d_{k} \mathbf{v}_{k} \end{gathered}

\mathbf{0}=\mathbf{v}-\mathbf{v}=\left(c_{1}-d_{1}\right) \mathbf{v}_{1}+\left(c_{2}-d_{2}\right) \mathbf{v}_{2}+\cdots+\left(c_{k}-d_{k}\right) \mathbf{v}_{k} .

\begin{aligned} c_{1}-d_{1} & =0, & c_{2}-d_{2} & =0, & \ldots, & c_{k}-d_{k} \\ c_{1} & =d_{1}, & c_{2} & =d_{2}, & \ldots, & c_{k} \end{aligned}=d_{k} .

\mathbf{v}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}

[\mathbf{v}]_{B} \stackrel{\text { def }}{=}\left(c_{1}, c_{2}, \ldots, c_{k}\right)

\mathbf{v}=c_{1} \mathbf{e}_{1}+c_{2} \mathbf{e}_{2}+\cdots+c_{n} \mathbf{e}_{n}

(5,1)=c_{1}(2,1)+c_{2}(1,2) .

\left[\begin{array}{cc|c} 2 & 1 & 5 \\ 1 & 2 & 1 \end{array}\right] \xrightarrow{R_{2}-\frac{1}{2} R_{1}}\left[\begin{array}{cc|c} 2 & 1 & 5 \\ 0 & 3 / 2 & -3 / 2 \end{array}\right] .

(5,4,3)=c_{1}(1,2,1)+c_{2}(2,1,1) .

\left[\begin{array}{ll|l} 1 & 2 & 5 \\ 2 & 1 & 4 \\ 1 & 1 & 3 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{3}-R_{1}}}\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -3 & -6 \\ 0 & -1 & -2 \end{array}\right] \xrightarrow{\substack{\frac{-1}{3} R_{2} \\ R_{3}+R_{2}}}\left[\begin{array}{ll|l} 1 & 2 & 5 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right] .

\mathbf{v}_{j}=0 \mathbf{v}_{1}+0 \mathbf{v}_{2}+\cdots+0 \mathbf{v}_{j-1}+1 \mathbf{v}_{j}+0 \mathbf{v}_{j+1}+\cdots+0 \mathbf{v}_{k} .

A=\left[\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 0 & 1 & -1 \\ 2 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right] .

\left[\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 0 & 1 & -1 \\ 2 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{4}-2 R_{1} \\ R_{5}-R_{1}}}\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_{3}-R_{2} \\ R_{4}-R_{2}}}\left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] .

B=\{(1,2,0,2,1),(0,1,1,1,0)\} .

(2,1,-3,1,2)=c_{1}(1,2,0,2,1)+c_{2}(0,1,1,1,0)

\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 2 & 1 & 1 \\ 0 & 1 & -3 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{4}-2 R_{1} \\ R_{5}-R_{1}}}\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_{3}-R_{2} \\ R_{4}-R_{2}}}\left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] .

\mathcal{S}=\operatorname{span}(B) \subset \mathbb{R}^{5},

2 \mathbf{v}-5 \mathbf{w}=2(2,1,-3,1,2)-5(1,0,-2,0,1)=(-1,2,4,2,-1) .

\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 2 & 1 & 2 \\ 0 & 1 & 4 \\ 2 & 1 & 2 \\ 1 & 0 & -1 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{4}-2 R_{1} \\ R_{5}-R_{1}}}\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 4 \\ 0 & 1 & 4 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_{3}-R_{2} \\ R_{4}-R_{2}}}\left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] .

2 \mathbf{v}-5 \mathbf{w}=(-1,2,4,2,-1)=-(1,2,0,2,1)+4(0,1,1,1,0),

\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & -2 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] \xrightarrow{\substack{R_{2}-2 R_{1} \\ R_{4}-2 R_{1} \\ R_{5}-R_{1}}}\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 1 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{array}\right] \xrightarrow{\substack{R_{3}-R_{2} \\ R_{4}-R_{2}}}\left[\begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] .

2[\mathbf{v}]_{B}-5[\mathbf{w}]_{B}=2(2,-3)-5(1,-2)=(-1,4),

(-1,1)=11 \mathbf{v}_{1}-12 \mathbf{v}_{2}, \quad \text { so } \quad[(-1,1)]_{B}=(11,-12) .

\left[\begin{array}{ll|l} 3 & 0 & 6 \\ 1 & 1 & 4 \end{array}\right] \xrightarrow{\frac{1}{3} R_{1}}\left[\begin{array}{ll|l} 1 & 0 & 2 \\ 1 & 1 & 4 \end{array}\right] \xrightarrow{R_{2}-R_{1}}\left[\begin{array}{ll|l} 1 & 0 & 2 \\ 0 & 1 & 2 \end{array}\right] .

P_{C \leftarrow B} \stackrel{\text { def }}{=}\left[\left[\mathbf{v}_{1}\right]_{C}\left|\left[\mathbf{v}_{2}\right]_{C}\right| \cdots \mid\left[\mathbf{v}_{k}\right]_{C}\right] .

\mathbf{v}_{1}=(1,1)=\frac{1}{3}(3,1)+\frac{2}{3}(0,1) \quad \text { and } \quad \mathbf{v}_{2}=(1,-1)=\frac{1}{3}(3,1)-\frac{4}{3}(0,1) .

\left[\mathbf{v}_{1}\right]_{C}=(1,2) / 3 \quad \text { and } \quad\left[\mathbf{v}_{2}\right]_{C}=(1,-4) / 3, \quad \text { so } \quad P_{C \leftarrow B}=\frac{1}{3}\left[\begin{array}{cc} 1 & 1 \\ 2 & -4 \end{array}\right] .

[\mathbf{v}]_{C}=P_{C \leftarrow B}[\mathbf{v}]_{B}=\frac{1}{3}\left[\begin{array}{cc} 1 & 1 \\ 2 & -4 \end{array}\right]\left[\begin{array}{l} 5 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 2 \end{array}\right],

\begin{aligned} P_{C \leftarrow B}[\mathbf{v}]_{B} & =\left[\left[\mathbf{v}_{1}\right]_{C}\left|\left[\mathbf{v}_{2}\right]_{C}\right| \cdots \mid\left[\mathbf{v}_{k}\right]_{C}\right]\left[\begin{array}{c} c_{1} \\ c_{2} \\ \vdots \\ c_{k} \end{array}\right] & & \left(\text { definition of } P_{C \leftarrow B}\right) \\ & =c_{1}\left[\mathbf{v}_{1}\right]_{C}+c_{2}\left[\mathbf{v}_{2}\right]_{C}+\cdots+c_{k}\left[\mathbf{v}_{k}\right]_{C} & & (\text { block matrix mult. }) \\ & =\left[c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}\right]_{C} & & (\text { by Theorem 3.1.2 }) \\ & =[\mathbf{v}]_{C} & & \left(\text { since }[\mathbf{v}]_{B}=\left(c_{1}, c_{2}, \ldots, c_{k}\right)\right) \end{aligned}

\left(P_{B \leftarrow C} P_{C \leftarrow B}\right) \mathbf{e}_{j}=P_{B \leftarrow C}\left(P_{C \leftarrow B}\left[\mathbf{v}_{j}\right]_{B}\right)=P_{B \leftarrow C}\left[\mathbf{v}_{j}\right]_{C}=\left[\mathbf{v}_{j}\right]_{B}=\mathbf{e}_{j} .

[\mathbf{w}]_{C}=P_{C \leftarrow B}[\mathbf{w}]_{B}=\frac{1}{3}\left[\begin{array}{cc} 1 & 1 \\ 2 & -4 \end{array}\right]\left[\begin{array}{c} -1 \\ 4 \end{array}\right]=\left[\begin{array}{c} 1 \\ -6 \end{array}\right] .

\begin{aligned} & (2,1,0,1,2) \\ & \quad=c_{1}(1,-1,1,0,2)+c_{2}(1,1,2,0,-1)+c_{3}(2,3,1,1,-1) \end{aligned}

\begin{aligned} & {\left[\begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ -1 & 1 & 3 & 1 \\ 1 & 2 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 2 & -1 & -1 & 2 \end{array}\right] \xrightarrow{\substack{R_{2}+R_{1} \\ R_{3}-R_{1} \\ R_{5}-2 R_{1}}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 0 & 2 & 5 & 3 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 1 & 1 \\ 0 & -3 & -5 & -2 \end{array}\right]} \\ & \xrightarrow{\substack{R_{2}-2 R_{3} \\ R_{5}+3 R_{3}}}\left[\begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 0 & 0 & 7 & 7 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & -8 & -8 \end{array}\right] \xrightarrow{\substack{R_{2}-7 R_{4} \\ R_{5}+8 R_{4}}}\left[\begin{array}{ccc|c} 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}

P_{C \leftarrow B}=\left[\left[\mathbf{v}_{1}\right]_{C}\left|\left[\mathbf{v}_{2}\right]_{C}\right|\left[\mathbf{v}_{3}\right]_{C}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 & -1 & -2 \\ 1 & 1 & 1 \end{array}\right] .

[\mathbf{v}]_{C}=P_{C \leftarrow B}[\mathbf{v}]_{B}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 & -1 & -2 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]=\left[\begin{array}{c} 1 \\ -9 \\ 6 \end{array}\right] .

[\mathbf{w}]_{C}=P_{C \leftarrow B}[\mathbf{w}]_{B}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 & -1 & -2 \\ 1 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 2 \\ 0 \\ -1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right] .

\begin{aligned} P_{C \leftarrow B} & =\left[\left[\mathbf{v}_{1}\right]_{C}\left|\left[\mathbf{v}_{2}\right]_{C}\right| \cdots \mid\left[\mathbf{v}_{k}\right]_{C}\right] \\ {[T] } & =\left[T\left(\mathbf{e}_{1}\right)\left|T\left(\mathbf{e}_{2}\right)\right| \cdots \mid T\left(\mathbf{e}_{k}\right)\right] . \end{aligned}

\mathbf{v}_{1}=(1,1,-1), \quad \mathbf{v}_{2}=(2,2,-1), \quad \text { and } \quad \mathbf{v}_{3}=(-1,-2,1) .

\left[\mathbf{v}_{1}\right]_{E}=(1,1,-1), \quad\left[\mathbf{v}_{2}\right]_{E}=(2,2,-1), \quad \text { and } \quad\left[\mathbf{v}_{3}\right]_{E}=(-1,-2,1) .

P_{E \leftarrow B}=\left[\left[\mathbf{v}_{1}\right]_{E}\left|\left[\mathbf{v}_{2}\right]_{E}\right|\left[\mathbf{v}_{3}\right]_{E}\right]=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 1 & 2 & -2 \\ -1 & -1 & 1 \end{array}\right] .

P_{B \leftarrow E}=P_{E \leftarrow B}^{-1}=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 1 & 2 & -2 \\ -1 & -1 & 1 \end{array}\right]^{-1}=\left[\begin{array}{ccc} 0 & -1 & -2 \\ 1 & 0 & 1 \\ 1 & -1 & 0 \end{array}\right] .

[T(\mathbf{v})]_{B}=[T]_{B}[\mathbf{v}]_{B} \quad \text { for all } \quad \mathbf{v} \in \mathbb{R}^{n} .

[T]_{B} \stackrel{\text { def }}{=}\left[\left[T\left(\mathbf{v}_{1}\right)\right]_{B}\left|\left[T\left(\mathbf{v}_{2}\right)\right]_{B}\right| \cdots \mid\left[T\left(\mathbf{v}_{n}\right)\right]_{B}\right] .

[T]_{B}=\left[\left[T\left(\mathbf{e}_{1}\right)\right]_{B} \mid\left[T\left(\mathbf{e}_{2}\right)\right]_{B}\right]=\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right] .

T(1,1)=(3,3) \quad \text { and } \quad T(1,-1)=(1,-1) .

\begin{aligned} (3,3)=3(1,1)+0(1,-1) & \text { so }[(3,3)]_{B}=(3,0), \quad \text { and } \\ (1,-1)=0(1,1)+1(1,-1) & \text { so }[(1,-1)]_{B}=(0,1) . \end{aligned}

[T]_{B}=\left[[T(1,1)]_{B} \mid[T(1,-1)]_{B}\right]=\left[\begin{array}{ll} 3 & 0 \\ 0 & 1 \end{array}\right] .

[T]_{C}=P_{C \leftarrow B}[T]_{B} P_{B \leftarrow C},

\begin{aligned} P_{C \leftarrow B}[T]_{B} P_{B \leftarrow C}[\mathbf{v}]_{C} & =P_{C \leftarrow B}[T]_{B}[\mathbf{v}]_{B} & & \left(P_{B \leftarrow C}[\mathbf{v}]_{C}=[\mathbf{v}]_{B}\right. \text { by Theorem 3.1.3 } \\ & =P_{C \leftarrow B}[T(\mathbf{v})]_{B} & & \left([T]_{B}[\mathbf{v}]_{B}=[T(\mathbf{v})]_{B}\right. \text { by Theorem 3. } \\ & =[T(\mathbf{v})]_{C} . & & \text { (by Theorem 3.1.3(a) again) } \end{aligned}

[T]=\left[\begin{array}{ccc} 2 & 1 & -3 \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right] .

P_{E \leftarrow B}=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 2 & -1 & 1 \\ 3 & 0 & 2 \end{array}\right], \quad \text { so } \quad P_{B \leftarrow E}=P_{E \leftarrow B}^{-1}=\frac{1}{7}\left[\begin{array}{ccc} 2 & 4 & -1 \\ 1 & -5 & 3 \\ -3 & -6 & 5 \end{array}\right] .

\begin{aligned} {[T]_{B} } & =P_{B \leftarrow E}[T] P_{E \leftarrow B} \\ & =\frac{1}{7}\left[\begin{array}{ccc} 2 & 4 & -1 \\ 1 & -5 & 3 \\ -3 & -6 & 5 \end{array}\right]\left[\begin{array}{ccc} 2 & 1 & -3 \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 2 & -1 & 1 \\ 3 & 0 & 2 \end{array}\right] \\ & =\frac{1}{7}\left[\begin{array}{ccc} 2 & 4 & -1 \\ 1 & -5 & 3 \\ -3 & -6 & 5 \end{array}\right]\left[\begin{array}{ccc} -5 & 3 & -7 \\ 4 & 2 & 1 \\ 8 & -1 & 5 \end{array}\right] \\ & =\frac{1}{7}\left[\begin{array}{ccc} -2 & 15 & -15 \\ -1 & -10 & 3 \\ 31 & -26 & 40 \end{array}\right] . \end{aligned}

B=P_{D \leftarrow C} A P_{D \leftarrow C}^{-1} .

A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cc} 5 & -1 \\ -2 & 0 \end{array}\right]

\begin{aligned} & P B P^{-1}=\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right]\left[\begin{array}{cc} 5 & -1 \\ -2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 / 2 & 1 / 2 \\ 1 / 2 & -1 / 2 \end{array}\right] \\ &= {\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ -1 & -1 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=A } \end{aligned}

\operatorname{rank}(A)=\operatorname{rank}\left(P B P^{-1}\right)=\operatorname{rank}(P B)=\operatorname{rank}(B) .

\left[\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right] \xrightarrow{R_{2}-2 R_{1}}\left[\begin{array}{ll} 1 & 2 \\ 0 & 0 \end{array}\right] \quad \text { and } \quad\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \xrightarrow{R_{2}-3 R_{1}}\left[\begin{array}{cc} 1 & 2 \\ 0 & -2 \end{array}\right] .

\operatorname{tr}(A) \stackrel{\text { def }}{=} a_{1,1}+a_{2,2}+\cdots+a_{n, n} .

\begin{aligned} & {[A B]_{i, i}=\sum_{j=1}^{n} a_{i, j} b_{j, i} \quad \text { for all } \quad 1 \leq i \leq m, \quad \text { and }} \\ & {[B A]_{j, j}=\sum_{i=1}^{m} b_{j, i} a_{i, j} \quad \text { for all } \quad 1 \leq j \leq n .} \end{aligned}

\begin{aligned} \operatorname{tr}(A B)=\sum_{i=1}^{m}[A B]_{i, i} & =\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i, j} b_{j, i} \\ & =\sum_{j=1}^{n} \sum_{i=1}^{m} b_{j, i} a_{i, j}=\sum_{j=1}^{n}[B A]_{j, j}=\operatorname{tr}(B A), \end{aligned}

\begin{aligned} & \operatorname{tr}(A B)=\operatorname{tr}\left(\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ -2 & 1 \end{array}\right]\right)=\operatorname{tr}\left(\left[\begin{array}{ll} -3 & 1 \\ -5 & 1 \end{array}\right]\right)=-2 \quad \text { and } \\ & \operatorname{tr}(B A)=\operatorname{tr}\left(\left[\begin{array}{cc} 1 & -1 \\ -2 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\right)=\operatorname{tr}\left(\left[\begin{array}{cc} -2 & -2 \\ 1 & 0 \end{array}\right]\right)=-2, \end{aligned}

\begin{gathered} \operatorname{tr}(A)=\operatorname{tr}\left(P B P^{-1}\right)=\operatorname{tr}\left(P\left(B P^{-1}\right)\right)=\underset{\uparrow}{\operatorname{tr}\left(\left(B P^{-1}\right) P\right)}=\operatorname{tr}\left(B\left(P^{-1} P\right)\right)=\operatorname{tr}(B) . \\ \text { by Theorem 3.1.6(c) } \end{gathered}

\operatorname{tr}(A B C)=\operatorname{tr}(A(B C))=\operatorname{tr}((B C) A)=\operatorname{tr}(B C A),

A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right], \quad B=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right], \quad \text { and } \quad C=\left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right],

\operatorname{tr}(A B C D)=\operatorname{tr}(B C D A)=\operatorname{tr}(C D A B)=\operatorname{tr}(D A B C) .

B=\{(3,1,4,-2,1),(2,3,-1,1,2),(-1,2,-2,4,-2)\}

\left(2 v_{2}+3 v_{3}, 3 v_{1}+v_{2}+3 v_{3}, 3 v_{1}+2 v_{2}+2 v_{3}\right) .

\left(4 v_{1}+v_{2}-v_{3}, 2 v_{1}-v_{2}-v_{3}, 2 v_{1}-v_{2}+v_{3}\right) .

\begin{aligned} B=\{ & (3,0,0,3),(3,0,1,1), \\ & (2,-1,0,1),(2,-1,3,-1)\}, \\ C= & \{(0,1,3,2),(-1,1,2,-1), \\ & (1,-1,-1,-1),(0,2,2,0)\} \end{aligned}

\begin{aligned} B=\{ & (-1,-1,1,0,2),(-1,-1,0,1,0), \\ C=\{ & (2,2,2,-1,-1),(1,1,1,2,0),(1,2,0,1,2)\}, \\ & (2,-1,3,0,1),(0,1,0,1,0),(3,-1,0),(1,1,0,1,0),(3,-1,2,3,3)\} \end{aligned}

A=\left[\mathbf{v}_{1}\left|\mathbf{v}_{2}\right| \cdots \mid \mathbf{v}_{n}\right] .

\frac{\text { volume of output region }}{\text { volume of input region }} .

\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B) \quad \text { for all } \quad A, B \in \mathcal{M}_{n} .

\operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| c \mathbf{a}_{j}|\cdots| \mathbf{a}_{n}\right]\right)=c \cdot \operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{a}_{j}|\cdots| \mathbf{a}_{n}\right]\right) .

\begin{aligned} & \operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{v}+\mathbf{w}|\cdots| \mathbf{a}_{n}\right]\right) \\ & \quad=\operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{v}|\cdots| \mathbf{a}_{n}\right]\right)+\operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{w}|\cdots| \mathbf{a}_{n}\right]\right) . \end{aligned}

\begin{aligned} & \operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{v}+c \mathbf{w}|\cdots| \mathbf{a}_{n}\right]\right) \\ & =\operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{v}|\cdots| \mathbf{a}_{n}\right]\right)+c \cdot \operatorname{det}\left(\left[\mathbf{a}_{1}|\cdots| \mathbf{w}|\cdots| \mathbf{a}_{n}\right]\right) . \end{aligned}