Rules, the transpose, and solving the system

(a) Four terms first. $$ \begin{aligned} &(\mathbf{A}+2\mathbf{I})(\mathbf{A}-3\mathbf{I})\\ &=\mathbf{A}^{2}-3\mathbf{A}\mathbf{I}\\ &\quad+2\mathbf{I}\mathbf{A}-6\mathbf{I}^{2}\,. \end{aligned} $$

(a) Then the identity. $\mathbf{A}\mathbf{I}=\mathbf{I}\mathbf{A}=\mathbf{A}$ and $\mathbf{I}^{2}=\mathbf{I},$ so the middle terms are $-3\mathbf{A}+2\mathbf{A}=-\mathbf{A}$ and $$ \begin{aligned} &(\mathbf{A}+2\mathbf{I})(\mathbf{A}-3\mathbf{I})\\ &=\mathbf{A}^{2}-\mathbf{A}-6\mathbf{I}\,. \end{aligned} $$

(b) Four terms, and they stay four. $$ \begin{aligned} (\mathbf{A}-\mathbf{B})^{2} &=\mathbf{A}^{2}-\mathbf{A}\mathbf{B}\\ &\quad-\mathbf{B}\mathbf{A}+\mathbf{B}^{2}\,. \end{aligned} $$ In (a) the middle terms are $\mathbf{A}\mathbf{I}$ and $\mathbf{I}\mathbf{A},$ which are equal by Fact 13.4: the identity commutes with every square matrix. In (b) they are $\mathbf{A}\mathbf{B}$ and $\mathbf{B}\mathbf{A},$ which in general differ.

(c) The check. $$ \begin{aligned} \mathbf{A}+2\mathbf{I}&=\begin{pmatrix}3&2\\3&2\end{pmatrix},\\ \mathbf{A}-3\mathbf{I}&=\begin{pmatrix}-2&2\\3&-3\end{pmatrix}, \end{aligned} $$ and their product is $\begin{pmatrix}0&0\\0&0\end{pmatrix}.$ The other side agrees: $\mathbf{A}^{2}=\begin{pmatrix}7&2\\3&6\end{pmatrix},$ so $\mathbf{A}^{2}-\mathbf{A}-6\mathbf{I}=\mathbf{0}.$

(c) What it shows. Neither $\mathbf{A}+2\mathbf{I}$ nor $\mathbf{A}-3\mathbf{I}$ is the zero matrix, yet their product is: failure (2) of Watch out 13.5. For numbers, $(a+2)(a-3)=0$ would force $a=-2$ or $a=3;$ here $\mathbf{A}$ is neither $-2\mathbf{I}$ nor $3\mathbf{I}.$

(a) Row by column. $$ \begin{aligned} \mathbf{N}^{2} &=\begin{pmatrix}4-4&-8+8\\2-2&-4+4\end{pmatrix}\\ &=\begin{pmatrix}0&0\\0&0\end{pmatrix}. \end{aligned} $$ Among real numbers only $0$ has $a^{2}=0;$ here $\mathbf{N}\ne\mathbf{0}.$

(b) Two different matrices. With $\mathbf{C}=\mathbf{I}$ and $\mathbf{D}=\mathbf{I}+\mathbf{N}=\begin{pmatrix}3&-4\\1&-1\end{pmatrix},$ $$ \begin{aligned} \mathbf{N}\mathbf{D}&=\mathbf{N}+\mathbf{N}^{2}=\mathbf{N}\,,\\ \mathbf{N}\mathbf{C}&=\mathbf{N}\mathbf{I}=\mathbf{N}\,. \end{aligned} $$ So $\mathbf{N}\mathbf{C}=\mathbf{N}\mathbf{D}$ although $\mathbf{C}\ne\mathbf{D}$ (and $\mathbf{N}\ne\mathbf{0}$).

(c) Failures (2) and (3). $\mathbf{N}\mathbf{N}=\mathbf{0}$ with neither factor zero is failure (2); $\mathbf{N}$ cannot be cancelled from $\mathbf{N}\mathbf{C}=\mathbf{N}\mathbf{D}$ is failure (3). Any $\mathbf{D}=\mathbf{I}+k\mathbf{N}$ works as well.

(a) The rule twice. With $\mathbf{P}=\mathbf{A}\mathbf{B}$ and $\mathbf{Q}=\mathbf{C},$ $$ \begin{aligned} (\mathbf{A}\mathbf{B}\mathbf{C})^{\top} &=\bigl((\mathbf{A}\mathbf{B})\mathbf{C}\bigr)^{\top}\\ &=\mathbf{C}^{\top}(\mathbf{A}\mathbf{B})^{\top}\\ &=\mathbf{C}^{\top}\mathbf{B}^{\top}\mathbf{A}^{\top}\,. \end{aligned} $$

(b) The orders. $\mathbf{A}\mathbf{B}\mathbf{C}$ is $(2\times3)(3\times4)(4\times1),$ that is $2\times1,$ and its transpose is $1\times2.$ The factors $\mathbf{C}^{\top},$ $\mathbf{B}^{\top},$ $\mathbf{A}^{\top}$ are $1\times4,$ $4\times3,$ $3\times2{:}$ they meet in the middle, and the ends give $1\times2$ again. The original order, $\mathbf{A}^{\top}\mathbf{B}^{\top}\mathbf{C}^{\top},$ would be $(3\times2)(4\times3)(1\times4),$ which does not exist.

(c) Symmetric. By Fact 13.9, $$ \begin{aligned} \mathbf{S}^{\top}&=\mathbf{A}^{\top}+(\mathbf{A}^{\top})^{\top}\\ &=\mathbf{A}^{\top}+\mathbf{A}=\mathbf{S}\,. \end{aligned} $$ For the given matrix, $$ \begin{aligned} \mathbf{S}&=\begin{pmatrix}1+1&4-2\\-2+4&3+3\end{pmatrix}\\ &=\begin{pmatrix}2&2\\2&6\end{pmatrix}, \end{aligned} $$ which is indeed unchanged by reflection in its main diagonal.

Swap, then clear the first column. Interchange rows $1$ and $2;$ then subtract $2\times$ row $1$ from row $3{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c} 0&2&3&1\\ 1&1&1&2\\ 2&3&-1&9 \end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&2&3&1\\ 2&3&-1&9 \end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&2&3&1\\ 0&1&-3&5 \end{array}\right) \end{aligned} $$

Second column. Interchange rows $2$ and $3,$ so that the leading entry is already $1;$ then subtract $2\times$ row $2$ from row $3{:}$ $$ \begin{aligned} \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&1&-3&5\\ 0&2&3&1 \end{array}\right)\\ \sim{}&\left(\begin{array}{ccc|c} 1&1&1&2\\ 0&1&-3&5\\ 0&0&9&-9 \end{array}\right) \end{aligned} $$

Back substitution. Divide row $3$ by $9{:}$ $z=-1.$ Row $2{:}$ $y=5+3z=2.$ Row $1{:}$ $x=2-y-z=2-2+1=1.$

Check. $2\cdot2+3\cdot(-1)=1,$ $1+2-1=2,$ $2+6+1=9.$ All three equations hold.

(a) Infinitely many. The last row is $0=0,$ so there is no contradiction. Leading entries in columns $1$ and $2;$ column $3$ has none, so $z$ is free. With $z=t{:}$ $y=3+t$ and $x=1-2t.$ All solutions: $(x,y,z)=(1-2t,\;3+t,\;t),$ $t\in\mathbb{R}.$

(b) None. The last row reads $0x+0y+0z=1,$ which no numbers satisfy. (The first two rows alone would have left $y$ free; that does not matter once one row is a contradiction.)

(c) Exactly one. No contradiction, and a leading entry in all three columns: $(x,y,z)=(-1,0,5).$

(a) Eliminate. Subtract $2\times$ row $1$ from row $2{:}$ $$ \begin{aligned} &\left(\begin{array}{cccc|c} 1&3&-1&2&4\\ 2&6&-1&5&11 \end{array}\right)\\ \sim{}&\left(\begin{array}{cccc|c} 1&3&-1&2&4\\ 0&0&1&1&3 \end{array}\right)\\ \sim{}&\left(\begin{array}{cccc|c} 1&3&0&3&7\\ 0&0&1&1&3 \end{array}\right) \end{aligned} $$ (the last step adds row $2$ to row $1,$ clearing above the leading entry).

(a) Read off. Leading entries in columns $1$ and $3;$ columns $2$ and $4$ have none, so $x_{2}=\alpha$ and $x_{4}=\beta$ are free: two degrees of freedom. The rows give $$ \begin{aligned} x_{3}&=3-\beta\,,\\ x_{1}&=7-3\alpha-3\beta\,. \end{aligned} $$

(b) The structure. Collect the constants, the $\alpha$ terms and the $\beta$ terms: $$ \begin{aligned} \begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix} &=\begin{pmatrix}7\\0\\3\\0\end{pmatrix} +\alpha\begin{pmatrix}-3\\1\\0\\0\end{pmatrix}\\ &\quad+\beta\begin{pmatrix}-3\\0\\-1\\1\end{pmatrix}. \end{aligned} $$

(b) The check. $\mathbf{s}=(7,0,3,0){:}$ $7-3=4$ and $14-3=11.$ $\mathbf{r}_{1}=(-3,1,0,0){:}$ $-3+3=0$ and $-6+6=0.$ $\mathbf{r}_{2}=(-3,0,-1,1){:}$ $-3+1+2=0$ and $-6+1+5=0.$ By Fact 13.23 the set is exactly $\mathbf{s}$ plus all homogeneous solutions; with two degrees of freedom it is a plane through $\mathbf{s},$ not a line.

(a) The difference. $\mathbf{h}=\mathbf{v}-\mathbf{u}=(1,-2,1),$ and $$ \mathbf{A}\mathbf{h}=\mathbf{A}\mathbf{v}-\mathbf{A}\mathbf{u} =\mathbf{b}-\mathbf{b}=\mathbf{0}\,. $$ It is not the zero vector, since $\mathbf{u}\ne\mathbf{v}.$

(b) A whole line of solutions. $$ \begin{aligned} \mathbf{A}(\mathbf{u}+\alpha\mathbf{h}) &=\mathbf{A}\mathbf{u}+\alpha\mathbf{A}\mathbf{h}\\ &=\mathbf{b}+\alpha\mathbf{0}=\mathbf{b}\,. \end{aligned} $$ For $\alpha=3{:}$ $\mathbf{u}+3\mathbf{h}=(4,-4,6).$

(c) Two give infinitely many. If a system has two different solutions $\mathbf{u}\ne\mathbf{v},$ then (b) gives one solution for every real $\alpha,$ and different $\alpha$ give different vectors because $\mathbf{h}\ne\mathbf{0}.$ So a second solution already forces infinitely many: after "none" and "exactly one", the only possibility left is "infinitely many", as Fact 13.20 says. This argument uses only linearity, not elimination.