Answer
Part A:
$\begin{bmatrix}
16\\
\end{bmatrix}$
Part B:
$\begin{bmatrix}
6 & 4 & 2 & 8\\
9 & 6 & 3 & 12\\
0 & 0 & 0 & 0\\
3 & 2 & 1 & 4\\
\end{bmatrix}$
Part C:
Not Possible.
Work Step by Step
A has 1 row and 4 columns (1 x 4).
B has 4 rows and 1 column (4 x 1).
Part A:
$\begin{bmatrix}
3(2) + 2(3) + 1(0) + 4(1)\\
\end{bmatrix}$ = $\begin{bmatrix}
6 + 6 + 0 + 4\\
\end{bmatrix}$ = $\begin{bmatrix}
16\\
\end{bmatrix}$
Part B:
$\begin{bmatrix}
2(3) & 2(2) & 2(1) & 2(4)\\
3(3) & 3(2) & 3(1) & 3(4)\\
0(3) & 0(2) & 0(1) & 0(4)\\
1(3) & 1(2) & 1(1) & 1(4)\\
\end{bmatrix}$ = $\begin{bmatrix}
6 & 4 & 2 & 8\\
9 & 6 & 3 & 12\\
0 & 0 & 0 & 0\\
3 & 2 & 1 & 4\\
\end{bmatrix}$
Part C:
Since the number of columns in A does not equal the number of rows in A, AA is not possible.
