Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 945: 34

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

According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. As $\begin{align} & D=\left| \begin{matrix} 2 & 2 & 3 \\ 4 & -1 & 1 \\ 5 & -2 & 6 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} 10 & 2 & 3 \\ -5 & -1 & 1 \\ 1 & -2 & 6 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 2 & 10 & 3 \\ 4 & -5 & 1 \\ 5 & 1 & 6 \\ \end{matrix} \right| \\ & {{D}_{z}}=\left| \begin{matrix} 2 & 2 & 10 \\ 4 & -1 & -5 \\ 5 & -2 & 1 \\ \end{matrix} \right| \end{align}$ Calculate the given four determinants. $\begin{align} & D=\left| \begin{matrix} 2 & 2 & 3 \\ 4 & -1 & 1 \\ 5 & -2 & 6 \\ \end{matrix} \right| \\ & =2\left( -6+2 \right)-2\left( 24-5 \right)+3\left( -8+5 \right) \\ & =-55 \end{align}$ $\begin{align} & {{D}_{x}}=\left| \begin{matrix} 10 & 2 & 3 \\ -5 & -1 & 1 \\ 1 & -2 & 6 \\ \end{matrix} \right| \\ & =10\left( -6+2 \right)-2\left( -30-1 \right)+3\left( 10+1 \right) \\ & =55 \end{align}$ $\begin{align} & {{D}_{y}}=\left| \begin{matrix} 2 & 10 & 3 \\ 4 & -5 & 1 \\ 5 & 1 & 6 \\ \end{matrix} \right| \\ & =2\left( -30-1 \right)-10\left( 24-5 \right)+3\left( 4+25 \right) \\ & =-165 \end{align}$ $\begin{align} & {{D}_{z}}=\left| \begin{matrix} 2 & 2 & 10 \\ 4 & -1 & -5 \\ 5 & -2 & 1 \\ \end{matrix} \right| \\ & =2\left( -1-10 \right)-2\left( 4+25 \right)+10\left( -8+5 \right) \\ & =-110 \end{align}$ Substitute the given values $\begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{55}{-55} \\ & =-1 \end{align}$ $\begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{-165}{-55} \\ & =3 \end{align}$ $\begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{-110}{-55} \\ & =2 \end{align}$ Therefore, $\left( x,y,z \right)=\left( -1,3,2 \right)$