## Elementary Linear Algebra 7th Edition

$$A^{-1}=\left[ \begin {array}{cccccc} -10&-4&27\\ 2&1&-5\\ -13&-5&35\end {array} \right].$$
To find $A^{-1}$, we have $$\left[ A \ \ I \right]= \left[ \begin {array}{cccccc} 10&5&-7&1&0&0\\ -5&1& 4&0&1&0\\ 3&2&-2&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[I \ \ A^{-1} \right]=\left[ \begin {array}{cccccc} 1&0&0&-10&-4&27\\ 0&1 &0&2&1&-5\\ 0&0&1&-13&-5&35\end {array} \right] .$$ Then $A^{-1}$ is given by $$A^{-1}=\left[ \begin {array}{cccccc} -10&-4&27\\ 2&1&-5\\ -13&-5&35\end {array} \right].$$