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.) $DH$
$
\begin{equation}
\begin{aligned}
DH =& \left[ \begin{array}{cc}
7 & 3
\end{array} \right] \left[ \begin{array}{cc}
3 & 1 \\
2 & -1
\end{array} \right]
&& \text{Substitute matrices $D$ and $H$}
\end{aligned}
\end{equation}
$
$
\begin{equation}
\begin{aligned}
& \text{Entry}
&& \text{Inner Product of}
&&& \text{Value}
&&&& \text{Matrix}
\\
\\
& C_{11}
&& \left[ \begin{array}{cc}
7 & 3
\end{array} \right]
\left[ \begin{array}{cc}
3 & 1 \\
2 & -1
\end{array} \right]
&&& 7 \cdot 3 + 3 \cdot 2 = 27
&&&& \left[ \begin{array}{cc}
27 &
\end{array} \right]
\\
\\
\\
\\
& C_{12}
&& \left[ \begin{array}{cc}
7 & 3
\end{array} \right]
\left[ \begin{array}{cc}
3 & 1 \\
2 & -1
\end{array} \right]
&&& 7 \cdot 1 + 3 \cdot (-1) = 4
&&&& \left[ \begin{array}{cc}
27 & 4
\end{array} \right]
\end{aligned}
\end{equation}
$
Thus, we have
$\displaystyle \left[ \begin{array}{cc}
7 & 3
\end{array} \right]
\left[ \begin{array}{cc}
3 & 1 \\
2 & -1
\end{array} \right]
=
\left[ \begin{array}{cc}
27 & 4
\end{array} \right]
$
b.) $HD$
$\displaystyle HD = \left[ \begin{array}{cc}
3 & 1 \\
2 & -1
\end{array} \right] \left[ \begin{array}{cc}
7 & 3
\end{array} \right]$
$HD$ is undefined because the number of columns of the first matrix must equal the number of rows of the second matrix.
Tuesday, October 31, 2017
College Algebra, Chapter 7, 7.2, Section 7.2, Problem 26
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