Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 945: 33

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

According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. As $\begin{align} & D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 3 & 2 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} 4 & 1 & 1 \\ 7 & -2 & 1 \\ 4 & 3 & 2 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 1 & 4 & 1 \\ 1 & 7 & 1 \\ 1 & 4 & 2 \\ \end{matrix} \right| \\ & {{D}_{z}}=\left| \begin{matrix} 1 & 1 & 4 \\ 1 & -2 & 7 \\ 1 & 3 & 4 \\ \end{matrix} \right| \end{align}$ Calculate the four determinants. $\begin{align} & D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 3 & 2 \\ \end{matrix} \right| \\ & =1\left( -4-3 \right)-1\left( 2-1 \right)+1\left( 3+2 \right) \\ & =-3 \end{align}$ $\begin{align} & {{D}_{x}}=\left| \begin{matrix} 4 & 1 & 1 \\ 7 & -2 & 1 \\ 4 & 3 & 2 \\ \end{matrix} \right| \\ & =4\left( -4-3 \right)-1\left( 14-4 \right)+1\left( 21+8 \right) \\ & =-9 \end{align}$ $\begin{align} & {{D}_{y}}=\left| \begin{matrix} 1 & 4 & 1 \\ 1 & 7 & 1 \\ 1 & 4 & 2 \\ \end{matrix} \right| \\ & =1\left( 14-4 \right)-1\left( 8-4 \right)+1\left( 4-7 \right) \\ & =3 \end{align}$ $\begin{align} & {{D}_{z}}=\left| \begin{matrix} 1 & 1 & 4 \\ 1 & -2 & 7 \\ 1 & 3 & 4 \\ \end{matrix} \right| \\ & =1\left( -8-21 \right)-1\left( 4-12 \right)+1\left( 7+8 \right) \\ & =-6 \end{align}$ Substitute the given values $\begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{-9}{-3} \\ & =3 \end{align}$ $\begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{3}{-3} \\ & =-1 \end{align}$ $\begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{-6}{-3} \\ & =2 \end{align}$ Therefore, $\left( x,y,z \right)=\left( 3,-1,2 \right)$