$\text{ If } A = \begin{bmatrix$}[r] 1 & 2 \ 3 & 4 \end{bmatrix}, B = \begin{bmatrix}[r] 2 & 1 \ 4 & 2 \end{bmatrix}, C = \begin{bmatrix}[r] -5 & 1 \ 7 & -4 \end{bmatrix}

Find:

(a) A + C

(b) B(A + C)

(c) 5B

(d) B(A + C) − 5B

(a) Given,

$\Rightarrow A + C = \begin{bmatrix$}[r] 1 & 2 \ 3 & 4 \end{bmatrix} + \begin{bmatrix}[r] -5 & 1 \ 7 & -4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + (-5) & 2 + 1 \ 3 + 7 & 4 + (-4) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -4 & 3 \ 10 & 0 \end{bmatrix}.

Hence, A + C = $\begin{bmatrix$}[r] -4 & 3 \ 10 & 0 \end{bmatrix}.

(b) From part (a), we get :

A + C = $\begin{bmatrix$}[r] -4 & 3 \ 10 & 0 \end{bmatrix}.

$\Rightarrow B(A + C) = \begin{bmatrix$}[r] 2 & 1 \ 4 & 2 \end{bmatrix} . \begin{bmatrix}[r] -4 & 3 \ 10 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] (2 \times -4 + 1 \times 10) & (2 \times 3 + 1 \times 0) \ (4 \times -4 + 2 \times 10) & (4 \times 3 + 2 \times 0) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -8 + 10 & 6 + 0 \ -16 + 20 & 12 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 6 \ 4 & 12 \end{bmatrix}.

Hence, B(A + C) = $\begin{bmatrix$}[r] 2 & 6 \ 4 & 12 \end{bmatrix}.

(c) Solving,

$5B = 5 . \begin{bmatrix$}[r] 2 & 1 \ 4 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 10 & 5 \ 20 & 10 \end{bmatrix}.

Hence, 5B = $\begin{bmatrix$}[r] 10 & 5 \ 20 & 10 \end{bmatrix}.

(d) From part b,

B(A + C) = $\begin{bmatrix$}[r] 2 & 6 \ 4 & 12 \end{bmatrix}.

From part c,

5B = $\begin{bmatrix$}[r] 10 & 5 \ 20 & 10 \end{bmatrix}.

Solving,

$\Rightarrow B(A + C) - 5B = \begin{bmatrix$}[r] 2 & 6 \ 4 & 12 \end{bmatrix} - \begin{bmatrix}[r] 10 & 5 \ 20 & 10 \end{bmatrix} \\[1em] \begin{bmatrix}[r] 2 - 10 & 6 - 5 \ 4 - 20 & 12 - 10 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -8 & 1 \ -16 & 2 \end{bmatrix}.

Hence, B(A + C) − 5B = $\begin{bmatrix$}[r] -8 & 1 \ -16 & 2 \end{bmatrix}.