Answer
a.
$AB=\left[\begin{array}{lll}
2 & -8 & 20\\
8 & 3 & 5\\
10 & 2 & 10
\end{array}\right]$
b.
$BA=\left[\begin{array}{ll}
12 & 14\\
9 & 3
\end{array}\right]$
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)
$c_{11}=2(3)+4(-1)=6-4=2$
$c_{12}=2(2)+4(-3)=4-12=-8$
and so on
$AB=\left[\begin{array}{lll}
6-4 & 4-12 & 0+20\\
9-1 & 6-3 & 0+5\\
12-2 & 8-6 & 0+10
\end{array}\right]$
$=\left[\begin{array}{lll}
2 & -8 & 20\\
8 & 3 & 5\\
10 & 2 & 10
\end{array}\right]$
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)
$d_{11}=3(2)+2(3)+0(4)=12$
$d_{12}=3(4)+2(1)+0(2)=14$
$d_{21}=-1(2)+(-3)(3)+5(4)=9$
$d_{22}=-1(4)+(-3)(1)+5(2)=3$
$BA=\left[\begin{array}{ll}
12 & 14\\
9 & 3
\end{array}\right]$
