## Precalculus (6th Edition) Blitzer

$\left( -5,-2,7 \right)$
According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. As \begin{align} & D=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ -1 & 3 & -1 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} 0 & 1 & 1 \\ -1 & -1 & 1 \\ -8 & 3 & -1 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 1 & 0 & 1 \\ 2 & -1 & 1 \\ -1 & -8 & -1 \\ \end{matrix} \right| \\ & {{D}_{z}}=\left| \begin{matrix} 1 & 1 & 0 \\ 2 & -1 & -1 \\ -1 & 3 & -8 \\ \end{matrix} \right| \end{align} Calculate the four determinants. \begin{align} & D=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ -1 & 3 & -1 \\ \end{matrix} \right| \\ & =1\left( 1-3 \right)-1\left( -2-\left( -1 \right) \right)+1\left( 6-1 \right) \\ & =4 \end{align} \begin{align} & {{D}_{x}}=\left| \begin{matrix} 0 & 1 & 1 \\ -1 & -1 & 1 \\ -8 & 3 & -1 \\ \end{matrix} \right| \\ & =0\left( 1-3 \right)+1\left( -1-\left( 3 \right) \right)-8\left( 1-\left( -1 \right) \right) \\ & =-20 \end{align} \begin{align} & {{D}_{y}}=\left| \begin{matrix} 1 & 0 & 1 \\ 2 & -1 & 1 \\ -1 & -8 & -1 \\ \end{matrix} \right| \\ & =1\left( 1-\left( -8 \right) \right)+1\left( -16-1 \right) \\ & =-8 \end{align} \begin{align} & {{D}_{z}}=\left| \begin{matrix} 1 & 1 & 0 \\ 2 & -1 & -1 \\ -1 & 3 & -8 \\ \end{matrix} \right| \\ & =1\left( 8-\left( -3 \right) \right)-2\left( -8 \right) \\ & =28 \end{align} Substitute the given values \begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{-20}{4} \\ & =-5 \end{align} \begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{-8}{4} \\ & =-2 \end{align} \begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{28}{4} \\ & =7 \end{align} Therefore, $\left( x,y,z \right)=\left( -5,-2,7 \right)$