College Algebra, Chapter 7, 7.2, Section 7.2, Problem 28

Suppose the matrices $A, B, C, D, E, F, G$ and $H$ are defined as



$
\begin{equation}
\begin{aligned}


A =& \left[ \begin{array}{cc}
2 & -5 \\
0 & 7
\end{array}
\right]

&& B = \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right]

&&& C = \left[ \begin{array}{ccc}
2 & \displaystyle \frac{-5}{2} & 0 \\
0 & 2 & -3
\end{array} \right]

&&&& D = \left[ \begin{array}{cc}
7 & 3
\end{array} \right]
\\
\\
\\
\\
E =& \left[ \begin{array}{c}
1 \\
2 \\
0
\end{array}
\right]

&& F = \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
\right]

&&& G = \left[ \begin{array}{ccc}
5 & -3 & 10 \\
6 & 1 & 0 \\
-5 & 2 & 2
\end{array} \right]

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


\end{aligned}
\end{equation}
$


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

a.) $BC$


$
\begin{equation}
\begin{aligned}

BC =& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
2 & \displaystyle \frac{-5}{2} & 0 \\
0 & 2 & -3
\end{array} \right]

\end{aligned}
\end{equation}
$


$BC$ is undefined because the number of columns of the first matrix must equal the number of rows of the second matrix.

b.) $BF$

$\displaystyle BF = \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]$


$
\begin{equation}
\begin{aligned}

& \text{Entry}
&& \text{Inner Product of}
&&& \text{Value}
&&&& \text{Matrix}
\\
\\
& C_{11}
&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]

&&& 3 \cdot 1 + \frac{1}{2} \cdot 0 + 5 \cdot 0 = 3
&&&& \left[ \begin{array}{ccc}
3 & & \\
& &
\end{array} \right]
\\
\\
\\
\\
& C_{12}
&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]

&&& 3 \cdot 0 + \frac{1}{2} \cdot 1 + 5 \cdot 0 = \frac{1}{2}
&&&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & \\
& &
\end{array} \right]

\\
\\
\\
\\
& C_{13}
&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]

&&& 3 \cdot 0 + \frac{1}{2} \cdot 0 + 5 \cdot 1 = 5
&&&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
& &
\end{array} \right]
\\
\\
\\
\\
& C_{21}
&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]
&&& 1 \cdot 1 + (-1) \cdot 0 + 3 \cdot 0 = 1
&&&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & &
\end{array} \right]
\\
\\
\\
\\
& C_{22}
&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]

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

\\
\\
\\
\\

& C_{23}
&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right] \left[ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right]
&&& 1 \cdot 0 + (-1) \cdot 0 + 3 \cdot 1 = 3
&&&& \left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right]

\end{aligned}
\end{equation}
$


Thus, we have


$
\begin{equation}
\begin{aligned}

\left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right]

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

=

\left[ \begin{array}{ccc}
3 & \displaystyle \frac{1}{2} & 5 \\
1 & -1 & 3
\end{array} \right]


\end{aligned}
\end{equation}
$