Answer
a.
$AB=\left[\begin{array}{lll}
6 & 8 & 16\\
11 & 16 & 24\\
1 & -1 & 12
\end{array}\right]$
b.
$BA=\left[\begin{array}{ll}
38 & 27\\
-16 & 4
\end{array}\right]$
Work Step by Step
A product of two matrices exists
if the first has as many columns
as the second matrix has rows.
The product of an $m\times\underline{n}$ matrix $A$ and an $\underline{n}\times p$ matrix $B$
is an $m\times p$ matrix $AB$.
a.
C=AB exists because
A is a $3\times\underline{2}$ matrix and B is a $\underline{2}\times 3$ matrix.
C is a 3$\times$3 matrix, $C=[c_{ij}]$,
where $c_{ij}$ is the (ith row of A) times (jth column of B)
$c_{11}=4(2)+2(-1)=8-2=6$
$c_{12}=4(3)+2(-2)=12-4=8$
and so on
$AB=\left[\begin{array}{lll}
8-2 & 12-4 & 16+0\\
12-1 & 18-2 & 24+0\\
6-5 & 9-10 & 12+0
\end{array}\right]$
$=\left[\begin{array}{lll}
6 & 8 & 16\\
11 & 16 & 24\\
1 & -1 & 12
\end{array}\right]$
b.
D=BA exists because
B is a $2\times\underline{3}$ matrix and A is a $\underline{3}\times 2$ matrix.
D is a $2\times 2$ matrix, $D=[d_{ij}]$,
where $d_{ij}$ is the (ith row of B) times (jth column of A)
$d_{11}=2(4)+3(6)+4(3)=8+18+12=38,$
$d_{12}=2(2)+3(1)+4(5)=4+3+20=27$
$d_{21}=-1(4)+(-2)(6)+0(3)=-4-12+0=-16$
$d_{22}=-1(2)+(-2)(1)+0(5)=-2-2+0=-4$
$BA=\left[\begin{array}{ll}
38 & 27\\
-16 & 4
\end{array}\right]$