Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 946: 37

$-200$

This is a fourth order determinant, so in order to calculate this determinant split this determinant of order 3. The second column has a maximum number of zeros, so expand this determinant with the second column as here: $\begin{align} & \left| \begin{matrix} 4 & 2 & 8 & -7 \\ -2 & 0 & 4 & 1 \\ 5 & 0 & 0 & 5 \\ 4 & 0 & 0 & -1 \\ \end{matrix} \right|={{\left( -1 \right)}^{1+2}}2\left| \begin{matrix} -2 & 4 & 1 \\ 5 & 0 & 5 \\ 4 & 0 & -1 \\ \end{matrix} \right| \\ & ={{\left( -1 \right)}^{3}}2\left| \begin{matrix} -2 & 4 & 1 \\ 5 & 0 & 5 \\ 4 & 0 & -1 \\ \end{matrix} \right| \\ & =-2\left| \begin{matrix} -2 & 4 & 1 \\ 5 & 0 & 5 \\ 4 & 0 & -1 \\ \end{matrix} \right| \end{align}$ Now calculate the third order determinant as follows: $\begin{align} & -2\left| \begin{matrix} -2 & 4 & 1 \\ 5 & 0 & 5 \\ 4 & 0 & -1 \\ \end{matrix} \right|=\left( -2 \right)\left\{ -2\left| \begin{matrix} 0 & 5 \\ 0 & -1 \\ \end{matrix} \right|-5\left| \begin{matrix} 4 & 1 \\ 0 & -1 \\ \end{matrix} \right|+4\left| \begin{matrix} 4 & 1 \\ 0 & 5 \\ \end{matrix} \right| \right\} \\ & =\left( -2 \right)\left\{ -2\left[ 0\left( -1 \right)-0\left( 5 \right) \right]-5\left[ 4\left( -1 \right)-0\left( 1 \right) \right]+4\left[ 4\left( 5 \right)-0\left( 1 \right) \right] \right\} \\ & =\left( -2 \right)\left\{ -2\left( 0 \right)-5\left( -4 \right)+4\left( 20 \right) \right\} \\ & =\left( -2 \right)\left\{ 0+20+80 \right\} \end{align}$ Simplify it to get, $\begin{align} & -2\left| \begin{matrix} -2 & 4 & 1 \\ 5 & 0 & 5 \\ 4 & 0 & -1 \\ \end{matrix} \right|=\left( -2 \right)\left\{ 0+20+80 \right\} \\ & =\left( -2 \right)\left( 100 \right) \\ & =-200 \end{align}$ Hence, $\left| \begin{matrix} 4 & 2 & 8 & -7 \\ -2 & 0 & 4 & 1 \\ 5 & 0 & 0 & 5 \\ 4 & 0 & 0 & -1 \\ \end{matrix} \right|=-200$