Q.If $A=\begin{pmatrix}6&-3\\4&2\end{pmatrix}$ and $B=\begin{pmatrix}-1&5\\3&-4\end{pmatrix}$, find

Inner product $2(1)+4(2)+6(3)=28$, returned as a $1\times 1$ matrix $[28]$.

The entries multiply pairwise and are added:

$2\times 1 + 4\times 2 + 6\times 3 = 2 + 8 + 18 = 28.$

A matrix times a matrix always produces a matrix, so this single number is reported as a $1\times 1$ matrix $[28]$. That is the distinction the examiner is testing between option (a) $28$ (a plain scalar) and option (d) $[28]$ (the correct matrix form).

$-3A$ scales each entry by $-3$ ⇒ option (a).

Scalar multiplication of a matrix multiplies every element by the scalar:

$-3A=-3\begin{bmatrix}2&3\\-1&4\end{bmatrix}=\begin{bmatrix}-3(2)&-3(3)\\-3(-1)&-3(4)\end{bmatrix}=\begin{bmatrix}-6&-9\\3&-12\end{bmatrix}.$

Element-wise addition ⇒ $\begin{bmatrix}9&8&11\\6&5&12\end{bmatrix}$.

Matrices of the same order are added entry by entry:

$\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}+\begin{bmatrix}-1&-2&1\\-4&-5&2\end{bmatrix}+\begin{bmatrix}9&8&7\\6&5&4\end{bmatrix}.$

Row 1: $(1-1+9,\;2-2+8,\;3+1+7)=(9,\;8,\;11)$.

Row 2: $(4-4+6,\;5-5+5,\;6+2+4)=(6,\;5,\;12)$.

$=\begin{bmatrix}9&8&11\\6&5&12\end{bmatrix}.$