Answer
The statement is false.
To make it true, replace the equals sign with "$\neq$".
Work Step by Step
If $A=\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]$, then $A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$.
---------
Counterexample:
Let A=$\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]$, B=$\left[\begin{array}{ll}
2 & 0\\
0 & 2
\end{array}\right]$
$A^{-1}=\displaystyle \frac{1}{1(1)-0}\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]$=$\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right],$
$B^{-1}=\displaystyle \frac{1}{2(2)-0}\left[\begin{array}{ll}
2 & 0\\
0 & 2
\end{array}\right]=\displaystyle \left[\begin{array}{ll}
1/2 & 0\\
0 & 1/2
\end{array}\right]$
$A^{-1}+B^{-1}=\left[\begin{array}{ll}
3/2 & 0\\
0 & 3/2
\end{array}\right]$
On the other hand,
$A+B=\left[\begin{array}{ll}
3 & 0\\
0 & 3
\end{array}\right], and$
$(A+B)^{-1}=\displaystyle \frac{1}{3(3)-0}\left[\begin{array}{ll}
3 & 0\\
0 & 3
\end{array}\right]=\displaystyle \left[\begin{array}{ll}
1/3 & 0\\
0 & 1/3
\end{array}\right]$
So, generally,
$(A+B)^{-1}\neq A^{-1}+B^{-1}$
The statement is false.
To make it true, replace the equals sign with "$\neq$".
