Answer
$AB=I_{2}$
$BA=I_{2}$
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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$a.$
$AB=\left[\begin{array}{ll}
4 & 1\\
7 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & -1\\
-7 & 4
\end{array}\right]$
$=\left[\begin{array}{ll}
4(2)+1(-7) & 4(-1)+1(4)\\
7(2)+2(-7) & 7(-1)+2(4)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$b.$
$BA=\left[\begin{array}{ll}
2 & -1\\
-7 & 4
\end{array}\right]\left[\begin{array}{ll}
4 & 1\\
7 & 2
\end{array}\right]$
$=\left[\begin{array}{ll}
2(4)+(-1)(7) & 2(1)+(-1)(2)\\
-7(4)+4(7) & -7(1)+4(2)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$AB=I_{2}$
$BA=I_{2}$
They are inverses of each other.
