\(\mathbf A=\begin{bmatrix}3&-1\\-10&9\end{bmatrix}\)
The inverse is given by
\(\mathbf A^{-1}=\dfrac1{\det\mathbf A}\mathrm{adj}\,\mathbf A\)
where \(\det\mathbf A\) is the determinant of the matrix, and \(\mathrm adj\,\mathbf A\) is the adjugate matrix, or transpose of the cofactor matrix.
\(\det\mathbf A=\begin{vmatrix}3&-1\\-10&9\end{vmatrix}=3(9)-(-1)(-10)=17\)
\(\mathrm{adj}\,\mathbf A=\begin{bmatrix}9&10\\1&3\end{bmatrix}^\top=\begin{bmatrix}9&1\\10&3\end{bmatrix}\)
So the inverse is
\(\mathbf A^{-1}=\dfrac1{17}\begin{bmatrix}9&1\\10&3\end{bmatrix}\)
