Answer
$X=\left[\begin{array}{rr}
0 & -5\\
-25 & -20\\
-10 & 10
\end{array}\right]$
Work Step by Step
Because $C$ on the RHS is a 3$\times$2 matrix,
for the equation to be defined, the LHS must also be a a 3$\times$2 matrix.
If X is a 3$\times$2 matrix, then so is $X+D$, as is $\displaystyle \frac{1}{5}(X+D).$
A solution exists.
$\displaystyle \frac{1}{5}(X+D)=C \qquad$ ... multiply with the scalar $5$
$ X+D=5C \qquad$ ... subtract D from both sides
$X=5C-D=5\left[\begin{array}{ll}
2 & 3\\
1 & 0\\
0 & 2
\end{array}\right]-\left[\begin{array}{ll}
10 & 20\\
30 & 20\\
10 & 0
\end{array}\right]$
$X=\left[\begin{array}{ll}
10-10 & 15-20\\
5-30 & 0-20\\
0-10 & 10-0
\end{array}\right]$
$X=\left[\begin{array}{ll}
0 & -5\\
-25 & -20\\
-10 & 10
\end{array}\right]$
