College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 34

Let

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

\qquad

D = \left[
\begin{array}{cc}
1 & 4 \\
0 & -1 \\
2 & 0
\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 $F(2C - D)$, or explain why it cannot be performed.



$
\begin{equation}
\begin{aligned}

F(2C - D) =& \left[ \begin{array}{ccc}
4 & 0 & 2 \\
-1 & 1 & 0 \\
7 & 5 & 0
\end{array} \right]
\left( 2 \left[ \begin{array}{cc}
\displaystyle \frac{1}{2} & 3 \\
2 & \displaystyle \frac{3}{2} \\
-2 & 1
\end{array} \right] - \left[ \begin{array}{cc}
1 & 4 \\
0 & -1 \\
2 & 0
\end{array} \right]
\right)
\\
\\
F(2C - D) =& \left[ \begin{array}{ccc}
4 & 0 & 2 \\
-1 & 1 & 0 \\
7 & 5 & 0
\end{array} \right]
\left( \left[ \begin{array}{cc}
1 & 6 \\
4 & 3 \\
-4 & 2
\end{array} \right] - \left[ \begin{array}{cc}
1 & 4 \\
0 & -1 \\
2 & 0
\end{array} \right]
\right)
\\
\\
F(2C - D) =& \left[ \begin{array}{ccc}
4 & 0 & 2 \\
-1 & 1 & 0 \\
7 & 5 & 0
\end{array} \right] \left[ \begin{array}{cc}
1-1 & 6-4 \\
4-0 & 3-(-1) \\
-4-2 & 2-0
\end{array} \right]
\\
\\
F(2C - D) =& \left[ \begin{array}{ccc}
4 & 0 & 2 \\
-1 & 1 & 0 \\
7 & 5 & 0
\end{array} \right] \left[ \begin{array}{cc}
0 & 2 \\
4 & 4 \\
-6 & 2
\end{array} \right]
\\
\\
F(2C - D) =& \left[ \begin{array}{cc}
4 \cdot 0 + 0 \cdot 4 + 2 \cdot (-6) & 4 \cdot 2 + 0 \cdot 4 + 2 \cdot 2 \\
-1 \cdot 0 + 1 \cdot 4 + 0 \cdot (-6) & -1 \cdot 2 + 1 \cdot 4 + 0 \cdot 2 \\
7 \cdot 0 + 5 \cdot 4 + 0 \cdot (-6) & 7 \cdot 2 + 5 \cdot 4 + 0 \cdot 2
\end{array} \right]
\\
\\
F(2C - D) =& \left[ \begin{array}{cc}
-12 & 12 \\
4 & 2 \\
20 & 34
\end{array} \right]

\end{aligned}
\end{equation}
$