Q.If $A=\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 0 \\ 2 & 5 \end{pmatrix}$, find $AB$ and $BA$, and hence verify that $AB\neq BA$.

$5\begin{bmatrix}0&1\\1&0\\1&1\end{bmatrix}-\begin{bmatrix}2&1\\3&-2\\1&3\end{bmatrix}=\begin{bmatrix}-2&4\\2&2\\4&2\end{bmatrix}$; 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 $3\times1$):

$\begin{bmatrix}-2&4\\2&2\\4&2\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}(-2)(2)+(4)(1)\\(2)(2)+(2)(1)\\(4)(2)+(2)(1)\end{bmatrix}=\begin{bmatrix}0\\6\\10\end{bmatrix}$

Step 4 — equate corresponding entries with $\begin{bmatrix}x-1\\y+1\\2z\end{bmatrix}$:

$x-1=0,\qquad y+1=6,\qquad 2z=10.$

Hence $x=1,\ y=5,\ z=5$.

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 $3\times1$):

$\begin{bmatrix} -2 & 4 \\ 2 & 2 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}=\begin{bmatrix} (-2)(2)+4(1) \\ 2(2)+2(1) \\ 4(2)+2(1) \end{bmatrix}=\begin{bmatrix} 0 \\ 6 \\ 10 \end{bmatrix}$.

Step 3 — equate to the right-hand side.

$\begin{bmatrix} 0 \\ 6 \\ 10 \end{bmatrix}=\begin{bmatrix} x+1 \\ y-1 \\ 3z \end{bmatrix}$.

Comparing entries: $x+1=0\Rightarrow x=-1$; $\ y-1=6\Rightarrow y=7$; $\ 3z=10\Rightarrow z=\dfrac{10}{3}$.

$2A + 3B = \begin{bmatrix} 19 & 22 \\ 24 & 15 \\ 1 & 1 \end{bmatrix}$; 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 $[-1,\ 1,\ 4]$:

$[-1,\ 1,\ 4]\begin{bmatrix} 19 & 22 \\ 24 & 15 \\ 1 & 1 \end{bmatrix} = \big[\,(-1)(19) + (1)(24) + (4)(1),\ \ (-1)(22) + (1)(15) + (4)(1)\,\big].$

Compute each entry:

$= [\,-19 + 24 + 4,\ \ -22 + 15 + 4\,] = [\,9,\ \ -3\,].$

Equating to $[x\ \ y]$ gives $x = 9,\ y = -3$.