$$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
2 & 2 & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7 \\
\end{pmatrix}
$$
Is this matrix invertible? I would like to show that it is invertible but first I should find the det(Matrix) which should not be equal to zero. To find the determinant, maybe the best idea is to use row operations and find an upper triangular of zeroes and then multiply the numbers on the diagonal to get the determinant. I have been doing some row operations and get this:
$$
\begin{pmatrix}
5 & 6 & 6 & 8 \\
0 & -1 & -4 & 1 \\
0 & 0 & 2 & 6 \\
-1 & 0 & 0 & -12 \\
\end{pmatrix}
$$
I just need to get rid of the -1 on the last row. But I am stuck. Thank you for your assistance.
Best Answer
A combination of Gaussian elimination and a well-known trick readily gives the answer. $$\det \begin{pmatrix} 5 & 6 & 6 & 8 \\ 2 & 2 & 2 & 8 \\ 6 & 6 & 2 & 8 \\ 2 & 3 & 6 & 7 \\ \end{pmatrix} =\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 2 & 2 & 2 & 6 \\ 6 & 6 & 2 & 2 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix} =4\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 3 & 3 & 1 & 1 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix} \\[0.5cm]=4\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 2 & 2 & 0 & -2 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix}=8\det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 1 & 1 & 0 & -1 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix} $$ and the last determinant is an odd integer, hence the original matrix is invertible. $$ \det\begin{pmatrix} 5 & 6 & 6 & 2 \\ 1 & 1 & 1 & 3 \\ 1 & 1 & 0 & -1 \\ 2 & 3 & 6 & 4 \\ \end{pmatrix}\equiv \det\begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix}\equiv 1\pmod{2}.$$