College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 30

Let

$\displaystyle C = \left[
\begin{array}{cc}
\displaystyle \frac{1}{2} & 3 \\
2 & \displaystyle \frac{3}{2} \\
-2 & 1
\end{array}
\right]
\qquad
B = \left[
\begin{array}{ccc}
1 & 2 & 4 \\
-2 & 1 & 0
\end{array}
\right]$

Carry out the indicated operation $CB$, or explain why it cannot be performed.

$\displaystyle CB = \left[
\begin{array}{cc}
\displaystyle \frac{1}{2} & 3 \\
2 & \displaystyle \frac{3}{2} \\
-2 & 1
\end{array}
\right]


\left[
\begin{array}{ccc}
1 & 2 & 4 \\
-2 & 1 & 0
\end{array}
\right]

=

\left[
\begin{array}{ccc}
\displaystyle \frac{1}{2} \cdot 1 + 3 \cdot (-2) & \displaystyle \frac{1}{2} \cdot 2 + 3 \cdot 1 & \displaystyle \frac{1}{2} \cdot 4 + 3 \cdot 2 \\
\displaystyle 2 \cdot 1 + \frac{3}{2} \cdot (-2) & \displaystyle 2 \cdot 2 + \frac{3}{2} \cdot 1 & \displaystyle 2 \cdot 4 + \frac{3}{2} \cdot 0 \\
-2 \cdot 1 + 1 \cdot (-2) & -2 \cdot 2 + 1 \cdot 1 & -2 \cdot 4 + 1 \cdot 0
\end{array}
\right]

=

\left[
\begin{array}{ccc}
\displaystyle \frac{-11}{2} & 4 & 2 \\
-1 & \displaystyle \frac{11}{2} & 8 \\
-4 & -3 & -8
\end{array}
\right]
$