Answer
$\left( A+B \right)\left( C-D \right)=\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right]$
Work Step by Step
First we will find the sum of $A\text{ and }B$ as below:
$\begin{align}
& A+B=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+\left[ \begin{matrix}
1 & 0 \\
0 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1+1 & 0+0 \\
0+0 & 1-1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & 0 \\
0 & 0 \\
\end{matrix} \right]
\end{align}$
Next we will find the difference between $C$ and $D$ as below:
$\begin{align}
& C-D=\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
-1 & 0 \\
0 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-1+1 & 0-0 \\
0-0 & 1+1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 0 \\
0 & 2 \\
\end{matrix} \right]
\end{align}$
Now consider,
$\begin{align}
& \left( A+B \right)\left( C-D \right)=\left[ \begin{matrix}
2 & 0 \\
0 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
0 & 0 \\
0 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2\left( 0 \right)+0\left( 0 \right) & 2\left( 0 \right)+0\left( 2 \right) \\
0\left( 0 \right)+0\left( 0 \right) & 0\left( 0 \right)+0\left( 2 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right]
\end{align}$
