Given,
The matrix $\mathrm{A}=\left[\begin{array}{ccc}2& -1& 4\\ -3& 0& 1\\ -1& 1& 2\end{array}\right]$

The determinant of A, |A| = 2(0 – 1) + 1( – 6 + 1) + 4( – 3) = – 19 ≠ 0

This indicates that the matrix have inverse

In order to find adjoint of the matrix A let's find the minors of the matrix A $=\left[\begin{array}{ccc}\left|\begin{array}{cc}0& 1\\ 1& 2\end{array}\right|& \left|\begin{array}{cc}-3& 1\\ -1& 2\end{array}\right|& \left|\begin{array}{cc}-3& 0\\ -1& 1\end{array}\right|\\ \left|\begin{array}{cc}-1& 4\\ 1& 2\end{array}\right|& \left|\begin{array}{cc}2& 4\\ -1& 2\end{array}\right|& \left|\begin{array}{cc}2& -1\\ -1& 1\end{array}\right|\\ \left|\begin{array}{cc}-1& 4\\ 0& 1\end{array}\right|& \left|\begin{array}{cc}2& 4\\ -3& 1\end{array}\right|& \left|\begin{array}{cc}2& -1\\ -3& 0\end{array}\right|\end{array}\right]=\left[\begin{array}{ccc}-1& -5& -3\\ -6& 8& 1\\ -1& 14& -3\end{array}\right]$

Co-factors

$=\left[\begin{array}{ccc}(-1{)}^{1+1}(-1)& (-1{)}^{1+2}(-5)& (-1{)}^{1+3}(-3)\\ (-1{)}^{2+1}(-6)& (-1{)}^{2+2}(8)& (-1{)}^{2+3}(1)\\ (-1{)}^{3+1}(-1)& (-1{)}^{3+2}(14)& (-1{)}^{3+3}(-3)\end{array}\right]=\left[\begin{array}{ccc}-1& 5& -3\\ 6& 8& -1\\ -1& -14& -3\end{array}\right]$

Transpose of the matrix A with co-factors

${{\mathrm{C}}_{\mathrm{A}}}^{\mathrm{T}}={\left[\begin{array}{ccc}-1& 5& -3\\ 6& 8& -1\\ -1& -14& -3\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{ccc}-1& 6& -1\\ 5& 8& -14\\ -3& -1& -3\end{array}\right]=\mathrm{adj}\mathrm{A}$

The inverse of the matrix $\mathrm{A},{\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\mathrm{adj}\mathrm{A}$

$=\frac{1}{-19}\left[\begin{array}{ccc}-1& 6& -1\\ 5& 8& -14\\ -3& -1& -3\end{array}\right]=\frac{1}{19}\left[\begin{array}{ccc}1& -6& 1\\ -5& -8& 14\\ 3& 1& 3\end{array}\right]$