Answer
$AC=\left[\begin{array}{rrr}{x-z}&{1-r}&{w-4}\\{2z}&{2r}&{8}\\{-2z}&{-2r}&{-8}\end{array}\right]$
Work Step by Step
If $A$ is an $m\times\boxed{n }$ matrix and $B$ is an $\boxed{n }\times k$ matrix,
then the product $AB$ is the $m\times k$ matrix whose $ij-$th entry is the product
of the (ith row in A) and (j-th column in B)
$(AB)_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+a_{i3}b_{3j}+\cdots+a_{in}b_{nj}$.
----
A is a 3$\times$2 matrix, $C$ is a 2$\times$3 matrix.
$AC$ is defined and is a 3$\times$3 matrix
$AC=\left[\begin{array}{cc}{1}&{-1}\\{0}&{2}\\{0}&{-2}\end{array}\right]\left[\begin{array}{ccc}{x}&{1}&{w}\\{z}&{r}&{4}\end{array}\right]$
$=\left[\begin{array}{rrr}{1 \cdot x+(-1) z}&{1\cdot 1+(-1)r}&{1\cdot w+(-1)\cdot 4}\\{0\cdot x+2z}&{0\cdot 1+2r}&{0\cdot w+2\cdot 4}\\{0\cdot x+(-2)z}&{0\cdot 1+(-2)r}&{0\cdot w+(-2)\cdot 4}\end{array}\right]$
$=\left[\begin{array}{ccc}{x-z}&{1-r}&{w-4}\\{2z}&{2r}&{8}\\{-2z}&{-2r}&{-8}\end{array}\right]$
