Answer
$$A^{-1}= \left[ \begin {array}{cccccccc} 27&-10&4&-29
\\ -16&5&-2&18\\ -
17&4&-2&20\\ -7&2&-1&8\end {array} \right]
.$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]=\left[ \begin {array}{cccccccc} 4&8&-7&14&1&0&0&0
\\ 2&5&-4&6&0&1&0&0\\ 0&2&1&-7&0&0
&1&0\\ 3&6&-5&10&0&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]= \left[ \begin {array}{cccccccc} 1&0&0&0&27&-10&4&-29
\\ 0&1&0&0&-16&5&-2&18\\ 0&0&1&0&-
17&4&-2&20\\ 0&0&0&1&-7&2&-1&8\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}= \left[ \begin {array}{cccccccc} 27&-10&4&-29
\\ -16&5&-2&18\\ -
17&4&-2&20\\ -7&2&-1&8\end {array} \right]
.$$
