## Precalculus (6th Edition) Blitzer

The determinant is $78$.
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} 3 & -1 & 1 & 2 \\ -2 & 0 & 0 & 0 \\ 2 & -1 & -2 & 3 \\ 1 & 4 & 2 & 3 \\ \end{matrix} \right|={{\left( -1 \right)}^{1+2}}\left( -1 \right)\left| \begin{matrix} -2 & 0 & 0 \\ 2 & -2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right|+{{\left( -1 \right)}^{3+2}}\left( -1 \right)\left| \begin{matrix} 3 & 1 & 2 \\ -2 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|+{{\left( -1 \right)}^{4+2}}4\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 0 & 0 \\ 2 & -2 & 3 \\ \end{matrix} \right| \\ & ={{\left( -1 \right)}^{3}}\left( -1 \right)\left| \begin{matrix} -2 & 0 & 0 \\ 2 & -2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right|+{{\left( -1 \right)}^{5}}\left( -1 \right)\left| \begin{matrix} 3 & 1 & 2 \\ -2 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|+{{\left( -1 \right)}^{6}}4\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 0 & 0 \\ 2 & -2 & 3 \\ \end{matrix} \right| \\ & =\left( -1 \right)\left( -1 \right)\left| \begin{matrix} -2 & 0 & 0 \\ 2 & -2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right|+\left( -1 \right)\left( -1 \right)\left| \begin{matrix} 3 & 1 & 2 \\ -2 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|+\left( 1 \right)4\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 0 & 0 \\ 2 & -2 & 3 \\ \end{matrix} \right| \\ & =\left| \begin{matrix} -2 & 0 & 0 \\ 2 & -2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right|+\left| \begin{matrix} 3 & 1 & 2 \\ -2 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|+4\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 0 & 0 \\ 2 & -2 & 3 \\ \end{matrix} \right| \end{align} Now calculate each third order determinant as here: First consider, \begin{align} & \left| \begin{matrix} -2 & 0 & 0 \\ 2 & -2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right|=-2\left| \begin{matrix} -2 & 3 \\ 2 & 3 \\ \end{matrix} \right|-2\left| \begin{matrix} 0 & 0 \\ 2 & 3 \\ \end{matrix} \right|+1\left| \begin{matrix} 0 & 0 \\ -2 & 3 \\ \end{matrix} \right| \\ & =-2\left[ -2\left( 3 \right)-2\left( 3 \right) \right]-2\left[ 0\left( 3 \right)-2\left( 0 \right) \right]+1\left[ 0\left( 3 \right)-\left( -2 \right)\left( 0 \right) \right] \\ & =-2\left[ -6-6 \right]-2\left( 0 \right)+1\left( 0 \right) \\ & =24 \end{align} Next consider the second determinant, \begin{align} & \left| \begin{matrix} 3 & 1 & 2 \\ -2 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|=3\left| \begin{matrix} 0 & 0 \\ 2 & 3 \\ \end{matrix} \right|-\left( -2 \right)\left| \begin{matrix} 1 & 2 \\ 2 & 3 \\ \end{matrix} \right|+1\left| \begin{matrix} 1 & 2 \\ 0 & 0 \\ \end{matrix} \right| \\ & =3\left[ 0\left( 3 \right)-2\left( 0 \right) \right]+2\left[ 1\left( 3 \right)-2\left( 2 \right) \right]+1\left[ 1\left( 0 \right)-0\left( 2 \right) \right] \\ & =3\left( 0 \right)+2\left( 3-4 \right)+1\left( 0 \right) \\ & =-2 \end{align} At last consider the third determinant, \begin{align} & 4\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 0 & 0 \\ 2 & -2 & 3 \\ \end{matrix} \right|=4\left( 3\left| \begin{matrix} 0 & 0 \\ -2 & 3 \\ \end{matrix} \right|-\left( -2 \right)\left| \begin{matrix} 1 & 2 \\ -2 & 3 \\ \end{matrix} \right|+2\left| \begin{matrix} 1 & 2 \\ 0 & 0 \\ \end{matrix} \right| \right) \\ & =4\left( 3\left[ 0\left( 3 \right)-\left( -2 \right)\left( 0 \right) \right]+2\left[ 1\left( 3 \right)-\left( -2 \right)\left( 2 \right) \right]+2\left[ 1\left( 0 \right)-0\left( 2 \right) \right] \right) \\ & =4\left( 3\left( 0 \right)+2\left( 3+4 \right)+2\left( 0 \right) \right) \\ & =56 \end{align} Now put these values in the original equations and simplify: \begin{align} & \left| \begin{matrix} 3 & -1 & 1 & 2 \\ -2 & 0 & 0 & 0 \\ 2 & -1 & -2 & 3 \\ 1 & 4 & 2 & 3 \\ \end{matrix} \right|=\left| \begin{matrix} -2 & 0 & 0 \\ 2 & -2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right|+\left| \begin{matrix} 3 & 1 & 2 \\ -2 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|+4\left| \begin{matrix} 3 & 1 & 2 \\ 2 & 0 & 0 \\ 2 & -2 & 3 \\ \end{matrix} \right| \\ & =24-2+56 \\ & =78 \end{align}