Given,
The matrix $\mathrm{A}=\left[\begin{array}{ccc}8& -4& 1\\ 10& 0& 6\\ 8& 1& 6\end{array}\right]$

The determinant of A, |A| = 8(0 – 6) + 4(60 – 48) + 1(10) = – 48 + 48 + 10 = 10 ≠ 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

$=\left[\begin{array}{ccc}\left|\begin{array}{cc}0& 6\\ 1& 6\end{array}\right|& \left|\begin{array}{cc}10& 6\\ 8& 6\end{array}\right|& \left|\begin{array}{cc}10& 0\\ 8& 1\end{array}\right|\\ \left|\begin{array}{cc}-4& 1\\ 1& 6\end{array}\right|& \left|\begin{array}{cc}8& 1\\ 8& 6\end{array}\right|& \left|\begin{array}{cc}8& -4\\ 8& 1\end{array}\right|\\ \left|\begin{array}{cc}-4& 1\\ 0& 6\end{array}\right|& \left|\begin{array}{cc}8& 1\\ 10& 6\end{array}\right|& \left|\begin{array}{cc}8& -4\\ 10& 0\end{array}\right|\end{array}\right]=\left[\begin{array}{ccc}-6& 12& 10\\ -25& 40& 40\\ -24& 38& 40\end{array}\right]$

Co-factors

$=\left[\begin{array}{ccc}(-1{)}^{1+1}(-6)& (-1{)}^{1+2}(12)& (-1{)}^{1+3}(10)\\ (-1{)}^{2+1}(-25)& (-1{)}^{2+2}(40)& (-1{)}^{2+3}(40)\\ (-1{)}^{3+1}(-24)& (-1{)}^{3+2}(38)& (-1{)}^{3+3}(40)\end{array}\right]=\left[\begin{array}{ccc}-6& -12& 10\\ 25& 40& -40\\ -24& -38& 40\end{array}\right]$

Transpose of the matrix A with co-factors

${\mathrm{C}}_{\mathrm{A}}^{\mathrm{T}}={\left[\begin{array}{ccc}-6& -12& 10\\ 25& 40& -40\\ -24& -38& 40\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{ccc}-6& 25& -24\\ -12& 40& -38\\ 10& -40& 40\end{array}\right]=\mathrm{adj}\mathrm{A}$

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

$=\frac{1}{10}\left[\begin{array}{ccc}-6& 25& -24\\ -12& 40& -38\\ 10& -40& 40\end{array}\right]$