Answer
$$A^{-1}=\left[ \begin {array}{cccccccc} -24&7&1&-2
\\ -10&3&0&-1\\ -
29&7&3&-2\\ 12&-3&-1&1\end {array} \right]
.$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]= \left[ \begin {array}{cccccccc} 1&-2&-1&-2&1&0&0&0
\\ 3&-5&-2&-3&0&1&0&0\\ 2&-5&-2&-5
&0&0&1&0\\ -1&4&4&11&0&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccccc} 1&0&0&0&-24&7&1&-2
\\ 0&1&0&0&-10&3&0&-1\\ 0&0&1&0&-
29&7&3&-2\\ 0&0&0&1&12&-3&-1&1\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}=\left[ \begin {array}{cccccccc} -24&7&1&-2
\\ -10&3&0&-1\\ -
29&7&3&-2\\ 12&-3&-1&1\end {array} \right]
.$$
