College Algebra, Chapter 7, 7.2, Section 7.2, Problem 12

Evaluate $ \left[ \begin{array}{ccc}
2 & 1 & 2 \\
6 & 3 & 4
\end{array} \right]
\left[\begin{array}{cc}
1 & -2 \\
3 & 6 \\
-2 & 0
\end{array} \right]$ or explain why it can not be performed.



$
\begin{equation}
\begin{aligned}

& \text{Entry} && \text{Inner Product of} &&& \text{Value} &&&& \text{Matrix}
\\
\\
& C_{11} && \left[ \begin{array}{ccc}
2 & 1 & 2 \\
6 & 3 & 4
\end{array} \right]

\left[ \begin{array}{cc}
1 & -2 \\
3 & 6 \\
-2 & 0
\end{array} \right]

&&& 2 \cdot 1 + 1 \cdot 3 + 2 \cdot (-2) = 1
&&&& \left[ \begin{array}{cc}
1 & \\
&
\end{array}
\right]

\\
\\
\\
\\

& C_{12} && \left[ \begin{array}{ccc}
2 & 1 & 2 \\
6 & 3 & 4
\end{array} \right]

\left[ \begin{array}{cc}
1 & -2 \\
3 & 6 \\
-2 & 0
\end{array} \right]

&&& 2 \cdot (-2) + 1 \cdot 6 + 2 \cdot 0 = 2
&&&& \left[ \begin{array}{cc}
1 & 2 \\
&
\end{array}
\right]

\\
\\
\\
\\

& C_{21}
&& \left[ \begin{array}{ccc}
2 & 1 & 2 \\
6 & 3 & 4
\end{array} \right]

\left[ \begin{array}{cc}
1 & -2 \\
3 & 6 \\
-2 & 0
\end{array} \right]

&&& 6 \cdot 1 + 0 \cdot 3 + 4 \cdot (-2) = -2
&&&& \left[ \begin{array}{cc}
1 & 2 \\
-2 &
\end{array}
\right]

\\
\\
\\
\\

& C_{22}
&& \left[ \begin{array}{ccc}
2 & 1 & 2 \\
6 & 3 & 4
\end{array} \right]

\left[ \begin{array}{cc}
1 & -2 \\
3 & 6 \\
-2 & 0
\end{array} \right]
&&& 6 \cdot (-2) + 0 \cdot 6 + 4 \cdot 0 = -12
&&&& \left[ \begin{array}{cc}
1 & 2 \\
-2 & -12
\end{array}
\right]


\end{aligned}
\end{equation}
$


Thus, we have

$\displaystyle \left[ \begin{array}{ccc}
2 & 1 & 2 \\
6 & 3 & 4
\end{array} \right]

\left[
\begin{array}{cc}
1 & -2 \\
3 & 6 \\
-2 & 0
\end{array}
\right]

=

\left[
\begin{array}{cc}
1 & 2 \\
-2 & -12
\end{array}
\right]

$