College Algebra, Chapter 7, 7.3, Section 7.3, Problem 4

Find the products $AB$ and $BA$ to verify that $B$ is the inverse of $A$ where $\displaystyle A = \left[ \begin{array}{cc}
2 & -3 \\
4 & -7
\end{array} \right]$ and $B = \left[ \begin{array}{cc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-3}{2} \\
2 & -1
\end{array} \right] $

We perform the matrix multiplications to show that $AB = I$ and $BA = I$


$
\begin{equation}
\begin{aligned}

AB =&
\left[ \begin{array}{cc}
2 & -3 \\
4 & -7
\end{array} \right]

\left[ \begin{array}{cc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-3}{2} \\
2 & 1
\end{array} \right]

=&

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

\\
\\
\\

BA =& \left[ \begin{array}{cc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-3}{2} \\
2 & 1
\end{array} \right]
\left[ \begin{array}{cc}
2 & -3 \\
4 & -7
\end{array} \right]

=&

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

=

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

\end{aligned}
\end{equation}
$


This shows that $A$ is the inverse of $B$, vice versa.