Answer
$\begin{bmatrix}
{-13}&{7}&{-12}\\
{-18}&{10}&{-14}\\
{17}&{-7}&{34}\end{bmatrix}$
Work Step by Step
In Problems 17 and 18, we found:
$CA=\begin{bmatrix}
{1}&{14}&{-14}\\
{2}&{22}&{-18}\\
{3}&{0}&{28}\end{bmatrix}$ and $CB=\begin{bmatrix}
{14}&{7}&{-2}\\
{20}&{12}&{-4}\\
{-14}&{7}&{-6}\end{bmatrix}$
Now, calculate $CA-CB=\begin{bmatrix}
{1-14}&{14-7}&{-14-(-2)}\\
{2-20}&{22-12}&{-18-(-4)}\\
{3-(-14)}&{0-7}&{28-(-6)}\end{bmatrix}\\=\begin{bmatrix}
{1-14}&{14-7}&{-14+2}\\
{2-20}&{22-12}&{-18+4}\\
{3+14}&{0-7}&{28+6}\end{bmatrix} \\=\begin{bmatrix}
{-13}&{7}&{-12}\\
{-18}&{10}&{-14}\\
{17}&{-7}&{34}\end{bmatrix}$
