Q.If $A$ is a matrix of order $2\times 3$ and $B$ is a matrix of order $3\times 4$, then the order of the product $AB$ is:

$5\begin{bmatrix}0&1\\1&0\\1&1\end{bmatrix}-\begin{bmatrix}2&1\\3&-2\\1&3\end{bmatrix}=\begin{bmatrix}-2&4\\2&2\\4&2\end{bmatrix}$; multiplying by $\begin{bmatrix}2\\1\end{bmatrix}$ gives $\begin{bmatrix}0\\6\\10\end{bmatrix}=\begin{bmatrix}x-1\\y+1\\2z\end{bmatrix}$, so $x=1,\ y=5,\ z=5$.

Step 1 — scalar multiply.

$5\begin{bmatrix}0&1\\1&0\\1&1\end{bmatrix}=\begin{bmatrix}0&5\\5&0\\5&5\end{bmatrix}$

Step 2 — subtract.

$\begin{bmatrix}0&5\\5&0\\5&5\end{bmatrix}-\begin{bmatrix}2&1\\3&-2\\1&3\end{bmatrix}=\begin{bmatrix}-2&4\\2&2\\4&2\end{bmatrix}$

Step 3 — multiply by the column $\begin{bmatrix}2\\1\end{bmatrix}$ (a $3\times2$ times a $2\times1$ gives a $3\times1$): …

Compute $5A-B$, multiply the resulting $3\times2$ matrix by $\begin{bmatrix}2\\1\end{bmatrix}$ to get $\begin{bmatrix}0\\6\\10\end{bmatrix}$, then equate to $\begin{bmatrix}x+1\\y-1\\3z\end{bmatrix}$.

Step 1 — scalar multiply and subtract.

$5\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 5 \\ 5 & 0 \\ 5 & 5 \end{bmatrix}$,

$5\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{bmatrix}-\begin{bmatrix} 2 & 1 \\ 3 & -2 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} -2 & 4 \\ 2 & 2 \\ 4 & 2 \end{bmatrix}$.

Step 2 — multiply by the column vector $\begin{bmatrix}2\\1\end{bmatrix}$ (a $3\times2$ times $2\times1$ gives $3\times1$):

…

$2A + 3B = \begin{bmatrix} 19 & 22 \\ 24 & 15 \\ 1 & 1 \end{bmatrix}$; then $[-1,1,4]$ times this is $[9\ \ {-3}]$, so $x = 9,\ y = -3$.

Step 1 — form $2A + 3B$:

$2\begin{bmatrix} 5 & 5 \\ 6 & 6 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 10 & 10 \\ 12 & 12 \\ -2 & 4 \end{bmatrix}, \qquad 3\begin{bmatrix} 3 & 4 \\ 4 & 1 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 9 & 12 \\ 12 & 3 \\ 3 & -3 \end{bmatrix}.$

Add:

$2A + 3B = \begin{bmatrix} 19 & 22 \\ 24 & 15 \\ 1 & 1 \end{bmatrix}.$

Step 2 — pre-multiply by $[-1,\ 1,\ 4]$: …