College Algebra, Chapter 7, 7.2, Section 7.2, Problem 32

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.) $(DA) B$


$
\begin{equation}
\begin{aligned}

(DA) B =& \left( \left[ \begin{array}{cc}
7 & 3 \end{array} \right]

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

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

\\
\\
\\

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

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

\\
\\
\\

=& \left[ \begin{array}{cc}
14 \cdot 3 + (-35) \cdot 1 & \displaystyle 14 \cdot \frac{1}{2} + (-35) \cdot (-1) & 14 \cdot 5 + (-35) \cdot 3
\end{array} \right]

\\
\\
\\

=& \left[ \begin{array}{ccc}
7 & 42 & -35
\end{array} \right]


\end{aligned}
\end{equation}
$


b.) $D(AB)$


$
\begin{equation}
\begin{aligned}

D(AB) =& \left[ \begin{array}{cc}
7 & 3
\end{array} \right]

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

\\
\\
\\

=& \left[ \begin{array}{cc}
7 & 3
\end{array} \right]

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

\\
\\
\\


=& \left[ \begin{array}{cc}
7 & 3
\end{array} \right]
\left[ \begin{array}{ccc}
1 & 6 & -5 \\
7 & -7 & 21
\end{array} \right]

\\
\\
\\

=& \left[ \begin{array}{ccc}
7 \cdot 1 + (-5 \cdot 7) & 7 \cdot 6 + (-5) \cdot (-7) & 7 \cdot (-5) + (-5) \cdot 21
\end{array} \right]

\\
\\
\\

=& \left[ \begin{array}{ccc}
-28 & 77 & -140
\end{array} \right]


\end{aligned}
\end{equation}
$

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