Answer
See the proof below.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c \\
\end{matrix} \right]$.
The inverse of matrix $ A=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right]$ is equal to ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$.
Now, we will compare the matrix to the original matrix, to get the inverse, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
Substitute the values to get,
$\begin{align}
& {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right] \\
& =\frac{1}{abc}\left[ \begin{matrix}
bc & 0 & 0 \\
0 & ca & 0 \\
0 & 0 & ab \\
\end{matrix} \right] \\
& {{A}^{-1}}=\left[ \begin{matrix}
\frac{1}{a} & 0 & 0 \\
0 & \frac{1}{b} & 0 \\
0 & 0 & \frac{1}{c} \\
\end{matrix} \right]
\end{align}$
Therefore, the inverse of the matrix ${{A}^{-1}}$ is $\left[ \begin{matrix}
\frac{1}{a} & 0 & 0 \\
0 & \frac{1}{b} & 0 \\
0 & 0 & \frac{1}{c} \\
\end{matrix} \right]$.
