Answer
See below
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
{2} & {32} & 1&4 \\
{26} & {104} & 26 & -13 \\
2 & 56 & 2 & 7\\
1 & 40 & 1 & 5 \end{array}\right] }\end{array} $$
So, we get
$$A=\begin{vmatrix} {2} & {32} & 1&4 \\
{26} & {104} & 26 & -13 \\
2 & 56 & 2 & 7\\
1 & 40 & 1 & 5
\end{vmatrix} \rightarrow \begin{vmatrix}
{1} & {40} & 1&5\\
{26} & {104} & 26 & -13 \\
2 & 56 & 2 & 7\\
2 & 32 & 1 & 4
\end{vmatrix} \rightarrow \begin{vmatrix}
{1} & {40} & 1& 5 \\
{0} & {-936} & 0 & -143 \\
0 & -24 & 0 & -3\\
0 & -48 & -1 & -6
\end{vmatrix} \rightarrow \begin{vmatrix}
{1} & {40} & 1& 5 \\
{0} & {0} & 0 & -26 \\
0 & -24 & 0 & -3\\
0 & 0 & -1 & 0
\end{vmatrix} \rightarrow \begin{vmatrix}
{1} & {40} & 1& 5 \\
{0} & {-24} & 0 & -3 \\
0 & 0 & 0 & -26\\
0 & 0 & -1 & 0
\end{vmatrix} \rightarrow \begin{vmatrix}
{1} & {40} & 1& 5 \\
{0} & {0} & 0 & -26 \\
0 & 0 & -1 & 0\\0 & -24 & 0 & -3
\end{vmatrix}$$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=-1 \cdot 1 \cdot (-24) \cdot (-1) \cdot (-26)=624}\end{array}
