$$\text{ If } A = \begin{bmatrix*}[r] 1 & 2 \\ 3 & 4 \end{bmatrix*}, B = \begin{bmatrix*}[r] 2 & 1 \\ 4 & 2 \end{bmatrix*}, C = \begin{bmatrix*}[r] -5 & 1 \\ 7 & -4 \end{bmatrix*}$$
Find:
(a) A + C
(b) B(A + C)
(c) 5B
(d) B(A + C) − 5B
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(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 = $$\begin{bmatrix*}[r] -8 & 1 \\ -16 & 2 \end{bmatrix*}$$.