Answer
$AB=I_{3}$
$BA=I_{3}$
They are inverses of each other.
Work Step by Step
If $A$ is an $n\times n$ matrix, then the inverse of $A$ is the $n\times n$ matrix $A^{-1}$ with the following properties:
$A^{-1}A=I_{n}$ and $AA^{-1}=I_{n}$
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$AB=\left[\begin{array}{lll}
3 & 2 & 4\\
1 & 1 & -6\\
2 & 1 & 12
\end{array}\right]\left[\begin{array}{lll}
9 & -10 & -8\\
-12 & 14 & 11\\
-1/2 & 1/2 & 1/2
\end{array}\right]$
$=\left[\begin{array}{lll}
27-12-1 & -30+28+1 & -24+22+2\\
9-12+3 & -10+14-3 & -8+11-3\\
18-12-6 & -20+14+6 & -16+11+6
\end{array}\right]$
$=\left[\begin{array}{lll}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]=I_{3}$
$BA=\left[\begin{array}{lll}
9 & -10 & -8\\
-12 & 14 & 11\\
-1/2 & 1/2 & 1/2
\end{array}\right]\left[\begin{array}{lll}
3 & 2 & 4\\
1 & 1 & -6\\
2 & 1 & 12
\end{array}\right]$
$=\left[\begin{array}{lll}
27-10-16 & 18-10-8 & 36+60-96\\
-36+14+22 & -24+14+11 & -48-84+132\\
-\frac{3}{2}+\frac{1}{2}+\frac{2}{2} & -1+\frac{1}{2}+\frac{1}{2} & -2-3+6
\end{array}\right]$
$=\left[\begin{array}{lll}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]=I_{3}$
$AB=I_{3}$
$BA=I_{3}$
They are inverses of each other.
