Answer
$$A^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ -
\frac{3}{4}&\frac{1}{4}&0\\ {\frac {7}{20}}&-\frac{1}{4}&\frac{1}{5}
\end {array} \right] .$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 3&4&0&0
&1&0\\ 2&5&5&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 0&1&0&-
\frac{3}{4}&\frac{1}{4}&0\\ 0&0&1&{\frac {7}{20}}&-\frac{1}{4}&\frac{1}{5}
\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ -
\frac{3}{4}&\frac{1}{4}&0\\ {\frac {7}{20}}&-\frac{1}{4}&\frac{1}{5}
\end {array} \right] .$$