Answer
$(A^{-1})^{-1}=A$
Work Step by Step
If $A=\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]$, then$ A$ is invertible if and only if $ad-bc\neq 0$,
and,
$A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$.
---------
$\displaystyle \frac{1}{ad-bc}=\frac{1}{3(4)-5(2)}=\frac{1}{2}$
$A^{-1}$ exists and
$B=A^{-1}=\displaystyle \frac{1}{2}\left[\begin{array}{ll}
4 & -5\\
-2 & 3
\end{array}\right]=\displaystyle \left[\begin{array}{ll}
2 & -5/2\\
-1 & 3/2
\end{array}\right]$
$B$ is the inverse of A.
Therefore
$BA=AB=I.$
We also read this as
"A is the inverse of B"
or
$B^{-1}=A$
or
$(A^{-1})^{-1}=A$