## Elementary Linear Algebra 7th Edition

$$\operatorname{adj}(A)=\left[ \begin {array}{cccc} 7&1&9&-13\\ 7&1&0&-4 \\ -4&2&-9&10\\ 2&-1&9&-5 \end {array} \right] .$$ $$A^{-1}=\frac{1}{9}\left[ \begin {array}{cccc} 7&1&9&-13\\ 7&1&0&-4 \\ -4&2&-9&10\\ 2&-1&9&-5 \end {array} \right].$$
The matrix is given by $A=\left[ \begin {array}{cccc} -1&2&0&1\\ 3&-1&4&1 \\ 0 &0&1&2\\ -1&1&1&2\end {array} \right] .$ To find $\operatorname{adj}(A)$, we calculate first the cofactor matrix of $A$ as follows $$\left[ \begin {array}{cccc} 7&7&-4&2\\ 1&1&2&-1 \\ 9&0&-9&9\\ -13&-4&10&-5 \end {array} \right] .$$ Now, the adjoint of $A$ is $$\operatorname{adj}(A)=\left[ \begin {array}{cccc} 7&1&9&-13\\ 7&1&0&-4 \\ -4&2&-9&10\\ 2&-1&9&-5 \end {array} \right] .$$ To find $A^{-1}$, we have to calculate $\det(A)$ which is given by $$\det(A)=9.$$ Finally, we have $$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)=\frac{1}{9}\left[ \begin {array}{cccc} 7&1&9&-13\\ 7&1&0&-4 \\ -4&2&-9&10\\ 2&-1&9&-5 \end {array} \right].$$