College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 36

Show that the matrices $\displaystyle A = \left[ \begin{array}{ccc}
2 & -1 & 3 \\
2 & -2 & 1 \\
0 & 1 & 1
\end{array} \right]$ and $\displaystyle B = \left[ \begin{array}{ccc}
\displaystyle \frac{-3}{2} & 2 & \displaystyle \frac{5}{2} \\
-1 & 1 & 2 \\
1 & -1 & -1
\end{array} \right]$ are inverses of each other by calculating the products $AB$ and $BA$

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


$
\begin{equation}
\begin{aligned}

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

=

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

=

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

\end{aligned}
\end{equation}
$


The matrices $AB$ and $BA$ both arrive in the same result so the matrices $A$ and $B$ are inverses of each other.