Answer
$\left[\begin{array}{ccc}
{6}&{14}&{-14}\\
{2}&{27}&{-18}\\
{3}&{0}&{33}\end{array}\right]$
Work Step by Step
$CA=\left[\begin{array}{ccc}
{1}&{14}&{-14}\\
{2}&{22}&{-18}\\
{3}&{0}&{28}\end{array}\right]\quad$(see sol. exercise 17)
$5I_{3}=5\cdot\left[\begin{array}{lll}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{lll}
5 & 0 & 0\\
0 & 5 & 0\\
0 & 0 & 5
\end{array}\right]$
the sum is defined, as both are 3 by 3 matrices.
$CA+5I_{3}=\left[\begin{array}{ccc}
{1+5}&{14}&{-14}\\
{2}&{22+5}&{-18}\\
{3}&{0}&{28+5}\end{array}\right]=\left[\begin{array}{ccc}
{6}&{14}&{-14}\\
{2}&{27}&{-18}\\
{3}&{0}&{33}\end{array}\right]$
