College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 32

Let

$\displaystyle C = \left[
\begin{array}{cc}
\displaystyle \frac{1}{2} & 3 \\
2 & \displaystyle \frac{3}{2} \\
-2 & 1
\end{array}
\right]
\qquad
F = \left[
\begin{array}{ccc}
4 & 0 & 2 \\
-1 & 1 & 0 \\
7 & 5 & 0
\end{array}
\right]$

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

$\displaystyle FC = \left[
\begin{array}{ccc}
4 & 0 & 2 \\
-1 & 1 & 0 \\
7 & 5 & 0
\end{array}
\right]

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

=

\left[
\begin{array}{cc}
\displaystyle 4 \cdot \frac{1}{2} + 0 \cdot 2 + 2 \cdot (-2) & \displaystyle 4 \cdot 3 + 0 \cdot \frac{3}{2} + 2 \cdot 1 \\
\displaystyle -1 \cdot \frac{1}{2} + 1 \cdot 2 + 0 \cdot (-2) & \displaystyle -1 \cdot 3 + 1 \cdot \frac{3}{2} + 0 \cdot 1 \\
\displaystyle 7 \cdot \frac{1}{2} + 5 \cdot 2 + 0 \cdot (-2) & \displaystyle 7 \cdot 3 + 5 \cdot \frac{3}{2} + 0 \cdot 1
\end{array}
\right]

=

\left[
\begin{array}{cc}
-2 & 14 \\
\displaystyle \frac{3}{2} & \displaystyle \frac{-3}{2} \\
\displaystyle \frac{27}{2} & \displaystyle \frac{57}{2}
\end{array}
\right]

$