Answer
a) $A+B=\left[ \begin{matrix}
6 & 4 \\
5 & 5 \\
\end{matrix} \right]$
b) $A-B=\left[ \begin{matrix}
-10 & 2 \\
-5 & -3 \\
\end{matrix} \right]$
c) $\left( -4 \right)A=\left[ \begin{matrix}
8 & -12 \\
0 & -4 \\
\end{matrix} \right]$
d) $3A+2B=\left[ \begin{matrix}
10 & 11 \\
10 & 11 \\
\end{matrix} \right]$
Work Step by Step
(a)
Perform the addition of the matrices $A$ and $B$ as follows:
$\begin{align}
& A+B=\left[ \begin{matrix}
-2 & 3 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
8 & 1 \\
5 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2+8 & 3+1 \\
0+5 & 1+4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6 & 4 \\
5 & 5 \\
\end{matrix} \right]
\end{align}$
Hence, $A+B=\left[ \begin{matrix}
6 & 4 \\
5 & 5 \\
\end{matrix} \right]$
(b)
Perform the subtraction of the matrices $A$ and $B$ as follows:
$\begin{align}
& A-B=\left[ \begin{matrix}
-2 & 3 \\
0 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
8 & 1 \\
5 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-2-8 & 3-1 \\
0-5 & 1-4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10 & 2 \\
-5 & -3 \\
\end{matrix} \right]
\end{align}$
Hence, $A-B=\left[ \begin{matrix}
-10 & 2 \\
-5 & -3 \\
\end{matrix} \right]$
(c)
Perform the multiplication of the matrices $A$ with constant number.
$\begin{align}
& \left( -4 \right)A=\left( -4 \right)\left[ \begin{matrix}
-2 & 3 \\
0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-4\times \left( -2 \right) & -4\times \left( 3 \right) \\
-4\times 0 & -4\times \left( 1 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
8 & -12 \\
0 & -4 \\
\end{matrix} \right]
\end{align}$
Hence, $\left( -4 \right)A=\left[ \begin{matrix}
8 & -12 \\
0 & -4 \\
\end{matrix} \right]$
(d)
Perform the multiplication with a content number and addition of the matrices $A$ and $B$ as follows:
$\begin{align}
& 3A+2B=3\left[ \begin{matrix}
-2 & 3 \\
0 & 1 \\
\end{matrix} \right]+2\left[ \begin{matrix}
8 & 1 \\
5 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3\times \left( -2 \right) & 3\times 3 \\
3\times 0 & 3\times 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
2\times 8 & 2\times 1 \\
2\times 5 & 2\times 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-6 & 9 \\
0 & 3 \\
\end{matrix} \right]+\left[ \begin{matrix}
16 & 2 \\
10 & 8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
10 & 11 \\
10 & 11 \\
\end{matrix} \right]
\end{align}$
Hence, $3A+2B=\left[ \begin{matrix}
10 & 11 \\
10 & 11 \\
\end{matrix} \right]$