Answer
$\left[\begin{array}{rrr}
{-2+x-z}&{2-r}&{w-6}\\
{10+2z}&{2r-2}&{10}\\
{-10-2z}&{2-2r}&{-10}\end{array}\right]$
Work Step by Step
By the Distributive Law,
$A(B+C)=AB+AC$,
and we have calculated AB and AC in previous exercises
$A(B+C)=\left[\begin{array}{ccc}
{-2}&{1}&{-2}\\
{10}&{-2}&{2}\\
{-10}&{2}&{-2}\end{array}\right] + \left[\begin{array}{ccc}
{x-z}&{1-r}&{w-4}\\
{2z}&{2r}&{8}\\
{-2z}&{-2r}&{-8}\end{array}\right]= \left[\begin{array}{ccc}
{-2+x-z}&{2-r}&{w-6}\\
{10+2z}&{2r-2}&{10}\\
{-10-2z}&{2-2r}&{-10}\end{array}\right]$
