Answer
$\begin{bmatrix}
15&21&-16\\
22&34&-22\\
-11&7&22
\end{bmatrix}$
Work Step by Step
To Find: $C(A+B)$
$\textbf {I}: A+B$
$\begin{bmatrix}
0&3&-5\\
1&2&6
\end{bmatrix}+\begin{bmatrix}
4&1&0\\
-2&3&-2
\end{bmatrix}$:
Let the resultant be matrix M.
We add two matrices by adding the respective individual elements of each matrix.
$\implies M = \begin{bmatrix}
4&4&-5\\
-1&5&4
\end{bmatrix}$
$\textbf{II}: CM$
$\begin{bmatrix}
4&1\\
6&2\\
-2&3
\end{bmatrix}\times\begin{bmatrix}
4&4&-5\\
-1&5&4
\end{bmatrix}$
Let the resultant matrix be $Z$.
$\implies Z_{1,1} = (4)(4)+(1)(-1) = 15$
$\implies Z_{1,2} = (4)(4)+(1)(5) = 21$
$\implies Z_{1,3} = (4)(-5)+(1)(4) = -16$
$\implies Z_{2,1} = (6)(4)+(2)(-1) = 22$
$\implies Z_{2,2} = (6)(4)+(2)(5) = 34$
$\implies Z_{2,3} = (6)(-5)+(2)(4) = -22$
$\implies Z_{3,1} = (-2)(4)+(3)(-1) = -11$
$\implies Z_{3,2} = (-2)(4)+(3)(5) = 7$
$\implies Z_{3,3} = (-2)(-5)+(3)(4) = 22$
$\therefore Z = \begin{bmatrix}
15&21&-16\\
22&34&-22\\
-11&7&22
\end{bmatrix}$
