College Algebra, Chapter 7, 7.3, Section 7.3, Problem 6

Find the products $AB$ and $BA$ to verify that $B$ is the inverse of $A$ where $\displaystyle A = \left[ \begin{array}{ccc}
3 & 2 & 4 \\
1 & 1 & -6 \\
2 & 1 & 12
\end{array} \right]$ and $B = \left[ \begin{array}{ccc}
9 & -10 & -8 \\
-12 & 14 & 11 \\
\displaystyle \frac{-1}{2} & \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}
\end{array} \right] $

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


$
\begin{equation}
\begin{aligned}

AB =&
\left[ \begin{array}{ccc}
3 & 2 & 4 \\
1 & 1 & -6 \\
2 & 1 & 12
\end{array} \right]
\left[ \begin{array}{ccc}
9 & -10 & -8 \\
-12 & 14 & 11 \\
\displaystyle \frac{-1}{2} & \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}
\end{array} \right]

=&

\left[ \begin{array}{ccc}
\displaystyle 3 \cdot 9 + 2 \cdot (-12) + 4 \cdot \left( \frac{-1}{2} \right) & \displaystyle 3 \cdot (-10) + 2 \cdot 14 + 4 \cdot \frac{1}{2} & 3 \cdot (-8) + 2 \cdot 11 + 4 \cdot \frac{1}{2} \\
\displaystyle 1 \cdot 9 + 1 \cdot (-12) + (-6) \cdot \left( \frac{-1}{2} \right) & \displaystyle 1 \cdot (-10) + 1 \cdot 14 + (-6) \cdot \frac{1}{2} & 1 \cdot (-8) + 1 \cdot 11 + (-6) \cdot \frac{1}{2} \\
\displaystyle 2 \cdot 9 + 1 \cdot (-12) + 12 \cdot \left( \frac{-1}{2} \right) & \displaystyle 2 \cdot (-10) + 1 \cdot 14 + 12 \cdot \frac{1}{2} & \displaystyle 2 \cdot (-8) + 1 \cdot 11 + 12 \cdot \frac{1}{2}
\end{array} \right]

\\
\\
\\

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

\\
\\
\\
\\

BA =& \left[ \begin{array}{ccc}
9 & -10 & -8 \\
-12 & 14 & 11 \\
\displaystyle \frac{-1}{2} & \displaystyle \frac{1}{2} & \displaystyle \frac{1}{2}
\end{array} \right]

\left[ \begin{array}{ccc}
3 & 2 & 4 \\
1 & 1 & -6 \\
2 & 1 & 12
\end{array} \right]

=&
\left[ \begin{array}{ccc}
9 \cdot 3 + (-10) \cdot 1 + (-8) \cdot 2 & 9 \cdot 2 + (-10) \cdot 1 + (-8) \cdot 1 & 9 \cdot 4 + (-10) \cdot (-6) + (-8) \cdot 12 \\
-12 \cdot 3 + 14 \cdot 1 + 11 \cdot 2 & -12 \cdot 2 + 14 \cdot 1 + 11 \cdot 1 & -12 \cdot 4 + 14 \cdot (-6) + 11 \cdot 12 \\
\displaystyle \frac{-1}{2} \cdot 3 + \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot 2 & \displaystyle \frac{-1}{2} \cdot 2 + \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot 1 & \displaystyle \frac{-1}{2} \cdot 4 + \frac{1}{2} \cdot (-6) + \frac{1}{2} \cdot 12
\end{array} \right]

\\
\\
\\

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



\end{aligned}
\end{equation}
$


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