## Precalculus (6th Edition) Blitzer

The determinant is $48$.
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} 1 & -3 & 2 & 0 \\ -3 & -1 & 0 & -2 \\ 2 & 1 & 3 & 1 \\ 2 & 0 & -2 & 0 \\ \end{matrix} \right|={{\left( -1 \right)}^{1+2}}\left( -3 \right)\left| \begin{matrix} -3 & 0 & -2 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+{{\left( -1 \right)}^{2+2}}\left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+{{\left( -1 \right)}^{3+2}}1\left| \begin{matrix} 1 & 2 & 0 \\ -3 & 0 & -2 \\ 2 & -2 & 0 \\ \end{matrix} \right| \\ & ={{\left( -1 \right)}^{3}}\left( -3 \right)\left| \begin{matrix} -3 & 0 & -2 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+{{\left( -1 \right)}^{4}}\left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+{{\left( -1 \right)}^{5}}1\left| \begin{matrix} 1 & 2 & 0 \\ -3 & 0 & -2 \\ 2 & -2 & 0 \\ \end{matrix} \right| \\ & =\left( -1 \right)\left( -3 \right)\left| \begin{matrix} -3 & 0 & -2 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+\left( 1 \right)\left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+\left( -1 \right)1\left| \begin{matrix} 1 & 2 & 0 \\ -3 & 0 & -2 \\ 2 & -2 & 0 \\ \end{matrix} \right| \\ & =3\left| \begin{matrix} -3 & 0 & -2 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+\left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+\left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ -3 & 0 & -2 \\ 2 & -2 & 0 \\ \end{matrix} \right| \end{align} Now calculate each third order determinant as follows: First consider, \begin{align} & 3\left| \begin{matrix} -3 & 0 & -2 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|=3\left( -3\left| \begin{matrix} 3 & 1 \\ -2 & 0 \\ \end{matrix} \right|-2\left| \begin{matrix} 0 & -2 \\ -2 & 0 \\ \end{matrix} \right|+2\left| \begin{matrix} 0 & -2 \\ 3 & 1 \\ \end{matrix} \right| \right) \\ & =3\left( -3\left[ 3\left( 0 \right)-\left( -2 \right)1 \right]-2\left[ 0\left( 0 \right)-\left( -2 \right)\left( -2 \right) \right]+2\left[ 0\left( 1 \right)-3\left( -2 \right) \right] \right) \\ & =3\left( \left( -3 \right)\left( 2 \right)-2\left( -4 \right)+2\left( 6 \right) \right) \\ & =42 \end{align} Next consider the second determinant, \begin{align} & \left| \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|=\left( -1 \right)\left( 1\left| \begin{matrix} 3 & 1 \\ -2 & 0 \\ \end{matrix} \right|-2\left| \begin{matrix} 2 & 0 \\ -2 & 0 \\ \end{matrix} \right|+2\left| \begin{matrix} 2 & 0 \\ 3 & 1 \\ \end{matrix} \right| \right) \\ & =\left( -1 \right)\left( 1\left[ 3\left( 0 \right)-\left( -2 \right)1 \right]-2\left[ 2\left( 0 \right)-\left( -2 \right)\left( 0 \right) \right]+2\left[ 2\left( 1 \right)-3\left( 0 \right) \right] \right) \\ & =\left( -1 \right)\left[ 1\left( 2 \right)-2\left( 0 \right)+2\left( 2 \right) \right] \\ & =-6 \end{align} At last consider the third determinant, \begin{align} & \left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ -3 & 0 & -2 \\ 2 & -2 & 0 \\ \end{matrix} \right|=\left( -1 \right)\left( 1\left| \begin{matrix} 0 & -2 \\ -2 & 0 \\ \end{matrix} \right|-\left( -3 \right)\left| \begin{matrix} 2 & 0 \\ -2 & 0 \\ \end{matrix} \right|+2\left| \begin{matrix} 2 & 0 \\ 0 & -2 \\ \end{matrix} \right| \right) \\ & =\left( -1 \right)\left( 1\left[ 0\left( 0 \right)-\left( -2 \right)\left( -2 \right) \right]-\left( -3 \right)\left[ 2\left( 0 \right)-\left( -2 \right)0 \right]+2\left[ 2\left( -2 \right)-0\left( 0 \right) \right] \right) \\ & =\left( -1 \right)\left[ -1\left( 4 \right)-\left( -3 \right)\left( 0 \right)+2\left( -4 \right) \right] \\ & =12 \end{align} Now put these values in the original equations and simplify: \begin{align} & \left| \begin{matrix} 1 & -3 & 2 & 0 \\ -3 & -1 & 0 & -2 \\ 2 & 1 & 3 & 1 \\ 2 & 0 & -2 & 0 \\ \end{matrix} \right|=3\left| \begin{matrix} -3 & 0 & -2 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+\left| \begin{matrix} 1 & 2 & 0 \\ 2 & 3 & 1 \\ 2 & -2 & 0 \\ \end{matrix} \right|+\left( -1 \right)\left| \begin{matrix} 1 & 2 & 0 \\ -3 & 0 & -2 \\ 2 & -2 & 0 \\ \end{matrix} \right| \\ & =42-6+12 \\ & =48 \end{align}