Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 945: 35

$\left( 2,3,1 \right)$

According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. As $\begin{align} & D=\left| \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} 4 & 0 & 2 \\ 5 & 2 & -1 \\ 13 & 3 & 0 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 1 & 4 & 2 \\ 0 & 5 & -1 \\ 2 & 13 & 0 \\ \end{matrix} \right| \\ & {{D}_{z}}=\left| \begin{matrix} 1 & 0 & 4 \\ 0 & 2 & 5 \\ 2 & 3 & 13 \\ \end{matrix} \right| \end{align}$ Calculate the four determinants. $\begin{align} & D=\left| \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & -1 \\ 2 & 3 & 0 \\ \end{matrix} \right| \\ & =1\left( 0+3 \right)+2\left( 0-4 \right) \\ & =-5 \end{align}$ $\begin{align} & {{D}_{x}}=\left| \begin{matrix} 4 & 0 & 2 \\ 5 & 2 & -1 \\ 13 & 3 & 0 \\ \end{matrix} \right| \\ & =4\left( 0+3 \right)+2\left( 15-26 \right) \\ & =-10 \end{align}$ $\begin{align} & {{D}_{y}}=\left| \begin{matrix} 1 & 4 & 2 \\ 0 & 5 & -1 \\ 2 & 13 & 0 \\ \end{matrix} \right| \\ & =1\left( 0+13 \right)+2\left( -4-10 \right) \\ & =-15 \end{align}$ $\begin{align} & {{D}_{z}}=\left| \begin{matrix} 1 & 0 & 4 \\ 0 & 2 & 5 \\ 2 & 3 & 13 \\ \end{matrix} \right| \\ & =1\left( 26-15 \right)+2\left( 0-8 \right) \\ & =-5 \end{align}$ Substitute the given values $\begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{-10}{-5} \\ & =2 \end{align}$ $\begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{-15}{-5} \\ & =3 \end{align}$ $\begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{-5}{-5} \\ & =1 \end{align}$ Therefore, $\left( x,y,z \right)=\left( 2,3,1 \right)$