Answer
$$A^{-1}=\left[ \begin {array}{cccccc} \frac{1}{2}&0&0\\
0&\frac{1}{3}&0\\ 0&0&\frac{1}{5}\end {array} \right].$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 2&0&0&1&0&0\\ 0&3&0&0
&1&0\\ 0&0&5&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]=\left[ \begin {array}{cccccc} 1&0&0&\frac{1}{2}&0&0\\ 0&1&0
&0&\frac{1}{3}&0\\0&0&1&0&0&\frac{1}{5}\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}=\left[ \begin {array}{cccccc} \frac{1}{2}&0&0\\
0&\frac{1}{3}&0\\ 0&0&\frac{1}{5}\end {array} \right].$$
