Answer
$$A^{-1}=\left[ \begin {array}{cccccc} -13&6&4\\
12&-5&-3\\ -5&2&1\end {array} \right].$$
Work Step by Step
To find $A^{-1}$, we have
$$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 1&2&2&1&0&0\\ 3&7&9&0
&1&0\\ -1&-4&-7&0&0&1\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[I \ \ A^{-1} \right]=\left[ \begin {array}{cccccc} 1&0&0&-13&6&4\\ 0&1&0
&12&-5&-3\\ 0&0&1&-5&2&1\end {array} \right]
.
$$
Then $A^{-1}$ is given by
$$A^{-1}=\left[ \begin {array}{cccccc} -13&6&4\\
12&-5&-3\\ -5&2&1\end {array} \right].$$