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(N/A) Let $A = \begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4 \end{

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(N/A) Let $A = \begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4 \end{bmatrix}$.First,calculate $AB$:$AB = \begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \ 1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4) \end{bmatrix}$$= \begin{bmatrix} 5 & 8 & 14 \ 0 & -1 & 1 \ -1 & 0 & 1 \end{bmatrix}$.Next,calculate $BA$:$BA = \begin{bmatrix} -1 & 1 & 0 \ 0 & -1 & 1 \ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 0 & 1 & 0 \ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \ 0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+(-1)(0)+1(0) \ 2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0) \end{bmatrix}$$= \begin{bmatrix} -1 & -1 & -3 \ 1 & 0 & 0 \ 6 & 11 & 6 \end{bmatrix}$.Since $AB 
e BA$,the given statement is proved.

Show that $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \ne \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$

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