Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 946: 39

The determinant is $195$.

This is a fourth order determinant, so in order to calculate this determinant, split it into a determinant of order 3. Expand along the second column as follows, $\begin{align} & \left| \begin{matrix} -2 & -3 & 3 & 5 \\ 1 & -4 & 0 & 0 \\ 1 & 2 & 2 & -3 \\ 2 & 0 & 1 & 1 \\ \end{matrix} \right|={{\left( -1 \right)}^{1+2}}\left( -3 \right)\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+{{\left( -1 \right)}^{2+2}}\left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+{{\left( -1 \right)}^{3+2}}2\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \\ \end{matrix} \right| \\ & ={{\left( -1 \right)}^{3}}\left( -3 \right)\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+{{\left( -1 \right)}^{4}}\left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+{{\left( -1 \right)}^{5}}2\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \\ \end{matrix} \right| \\ & =\left( -1 \right)\left( -3 \right)\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+\left( 1 \right)\left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+\left( -1 \right)2\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \\ \end{matrix} \right| \\ & =3\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+\left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+\left( -2 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \\ \end{matrix} \right| \end{align}$ Now calculate each third order determinant as follows: First consider, $\begin{align} & 3\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|=3\left( 1\left| \begin{matrix} 2 & -3 \\ 1 & 1 \\ \end{matrix} \right|-1\left| \begin{matrix} 0 & 0 \\ 1 & 1 \\ \end{matrix} \right|+2\left| \begin{matrix} 0 & 0 \\ 2 & -3 \\ \end{matrix} \right| \right) \\ & =3\left( 1\left[ 2\left( 1 \right)-1\left( -3 \right) \right]-1\left[ 0\left( 1 \right)-1\left( 0 \right) \right]+2\left[ 0\left( -3 \right)-2\left( 0 \right) \right] \right) \\ & =3\left( 1\left[ 2+3 \right]-1\left( 0 \right)+2\left( 0 \right) \right) \\ & =15 \end{align}$ Next consider the second determinant, $\begin{align} & \left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|=\left( -4 \right)\left( \left( -2 \right)\left| \begin{matrix} 2 & -3 \\ 1 & 1 \\ \end{matrix} \right|-1\left| \begin{matrix} 3 & 5 \\ 1 & 1 \\ \end{matrix} \right|+2\left| \begin{matrix} 3 & 5 \\ 2 & -3 \\ \end{matrix} \right| \right) \\ & =\left( -4 \right)\left( \left( -2 \right)\left[ 2\left( 1 \right)-1\left( -3 \right) \right]-1\left[ 3\left( 1 \right)-1\left( 5 \right) \right]+2\left[ 3\left( -3 \right)-2\left( 5 \right) \right] \right) \\ & =\left( -4 \right)\left( -2\left[ 2+3 \right]-1\left[ 3-5 \right]+2\left[ -9-10 \right] \right) \\ & =\left( -4 \right)\left( -10+2-38 \right) \end{align}$ Simplify this to get, $\begin{align} & \left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|=\left( -4 \right)\left( -10+2-38 \right) \\ & =\left( -4 \right)\left( -46 \right) \\ & =184 \end{align}$ At last consider the third determinant, $\begin{align} & \left( -2 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \\ \end{matrix} \right|=\left( -2 \right)\left( \left( -2 \right)\left| \begin{matrix} 0 & 0 \\ 1 & 1 \\ \end{matrix} \right|-1\left| \begin{matrix} 3 & 5 \\ 1 & 1 \\ \end{matrix} \right|+2\left| \begin{matrix} 3 & 5 \\ 0 & 0 \\ \end{matrix} \right| \right) \\ & =\left( -2 \right)\left( \left( -2 \right)\left[ 0\left( 1 \right)-1\left( 0 \right) \right]-1\left[ 3\left( 1 \right)-1\left( 5 \right) \right]+2\left[ 3\left( 0 \right)-0\left( 5 \right) \right] \right) \\ & =\left( -2 \right)\left( \left( -2 \right)\left( 0 \right)-1\left( 3-5 \right)+2\left( 0 \right) \right) \\ & =-4 \end{align}$ Now put these values in the original equations and simplify: $\begin{align} & \left| \begin{matrix} -2 & -3 & 3 & 5 \\ 1 & -4 & 0 & 0 \\ 1 & 2 & 2 & -3 \\ 2 & 0 & 1 & 1 \\ \end{matrix} \right|=3\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+\left( -4 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 2 & -3 \\ 2 & 1 & 1 \\ \end{matrix} \right|+\left( -2 \right)\left| \begin{matrix} -2 & 3 & 5 \\ 1 & 0 & 0 \\ 2 & 1 & 1 \\ \end{matrix} \right| \\ & =15+184-4 \\ & =195 \end{align}$