Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 945: 36

$\left( 0,4,2 \right)$

According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. As $\begin{align} & D=\left| \begin{matrix} 3 & 0 & 2 \\ 5 & -1 & 0 \\ 0 & 4 & 3 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} 4 & 0 & 2 \\ -4 & -1 & 0 \\ 22 & 4 & 3 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 3 & 4 & 2 \\ 5 & -4 & 0 \\ 0 & 22 & 3 \\ \end{matrix} \right| \\ & {{D}_{z}}=\left| \begin{matrix} 3 & 0 & 4 \\ 5 & -1 & -4 \\ 0 & 4 & 22 \\ \end{matrix} \right| \end{align}$ Calculate the four determinants. $\begin{align} & D=\left| \begin{matrix} 3 & 0 & 2 \\ 5 & -1 & 0 \\ 0 & 4 & 3 \\ \end{matrix} \right| \\ & =3\left( -3-0 \right)+2\left( 20-0 \right) \\ & =31 \end{align}$ $\begin{align} & {{D}_{x}}=\left| \begin{matrix} 4 & 0 & 2 \\ -4 & -1 & 0 \\ 22 & 4 & 3 \\ \end{matrix} \right| \\ & =4\left( -3+0 \right)+2\left( -16+22 \right) \\ & =0 \end{align}$ $\begin{align} & {{D}_{y}}=\left| \begin{matrix} 3 & 4 & 2 \\ 5 & -4 & 0 \\ 0 & 22 & 3 \\ \end{matrix} \right| \\ & =2\left( 110 \right)+3\left( -12-20 \right) \\ & =124 \end{align}$ $\begin{align} & {{D}_{z}}=\left| \begin{matrix} 3 & 0 & 4 \\ 5 & -1 & -4 \\ 0 & 4 & 22 \\ \end{matrix} \right| \\ & =3\left( -22+16 \right)+4\left( 20 \right) \\ & =62 \end{align}$ Substitute the given values $\begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{0}{31} \\ & =0 \end{align}$ $\begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{124}{31} \\ & =4 \end{align}$ $\begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{62}{31} \\ & =2 \end{align}$ Therefore, $\left( x,y,z \right)=\left( 0,4,2 \right)$