Show that if
$$
A=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{llll}
b_{11} & b_{12} & b_{13} & b_{14} \\
b_{21} & b_{22} & b_{23} & b_{24} \\
b_{31} & b_{32} & b_{33} & b_{34}
\end{array}\right] \text {, }
$$
then
$$
A B=\left[\begin{array}{lll}
a_{11} & 0 & 0 \\
a_{21} & 0 & 0
\end{array}\right] B+\left[\begin{array}{lll}
0 & a_{12} & 0 \\
0 & a_{22} & 0
\end{array}\right] B+\left[\begin{array}{lll}
0 & 0 & a_{13} \\
0 & 0 & a_{23}
\end{array}\right] B
$$
and hence that
$$
A B=\left[\begin{array}{l}
a_{11} \\
a_{21}
\end{array}\right]\left[\begin{array}{lll}
b_{11} & \cdots & b_{14}
\end{array}\right]+\left[\begin{array}{l}
a_{12} \\
a_{22}
\end{array}\right]\left[\begin{array}{llll}
b_{21} & b_{22} & b_{23} & b_{24}
\end{array}\right]+\left[\begin{array}{l}
a_{13} \\
a_{23}
\end{array}\right]\left[\begin{array}{lll}
b_{31} & \cdots & b_{34}
\end{array}\right] \text {. }
$$