Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 945: 31

$\left( 2,-3,4 \right)$

According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. As $\begin{align} & D=\left| \begin{matrix} 4 & -5 & -6 \\ 1 & -2 & -5 \\ 2 & -1 & 0 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} -1 & -5 & -6 \\ -12 & -2 & -5 \\ 7 & -1 & 0 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 4 & -1 & -6 \\ 1 & -12 & -5 \\ 2 & 7 & 0 \\ \end{matrix} \right| \\ & {{D}_{z}}=\left| \begin{matrix} 4 & -5 & -1 \\ 1 & -2 & -12 \\ 2 & -1 & 7 \\ \end{matrix} \right| \end{align}$ Calculate the four determinants. $\begin{align} & D=\left| \begin{matrix} 4 & -5 & -6 \\ 1 & -2 & -5 \\ 2 & -1 & 0 \\ \end{matrix} \right| \\ & =4\left( 0-5 \right)+5\left( 0+10 \right)-6\left( -1+4 \right) \\ & =12 \end{align}$ $\begin{align} & {{D}_{x}}=\left| \begin{matrix} -1 & -5 & -6 \\ -12 & -2 & -5 \\ 7 & -1 & 0 \\ \end{matrix} \right| \\ & =-1\left( 0-5 \right)+5\left( 0+35 \right)-6\left( 12+14 \right) \\ & =24 \end{align}$ $\begin{align} & {{D}_{y}}=\left| \begin{matrix} 4 & -1 & -6 \\ 1 & -12 & -5 \\ 2 & 7 & 0 \\ \end{matrix} \right| \\ & =4\left( 0+35 \right)+1\left( 0+10 \right)-6\left( 7+24 \right) \\ & =-36 \end{align}$ $\begin{align} & {{D}_{z}}=\left| \begin{matrix} 4 & -5 & -1 \\ 1 & -2 & -12 \\ 2 & -1 & 7 \\ \end{matrix} \right| \\ & =4\left( -14-12 \right)+5\left( 7+24 \right)-1\left( -1+4 \right) \\ & =48 \end{align}$ Substitute the given values $\begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{24}{12} \\ & =2 \end{align}$ $\begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{-36}{12} \\ & =-3 \end{align}$ $\begin{align} & z=\frac{{{D}_{z}}}{D} \\ & =\frac{48}{12} \\ & =4 \end{align}$ Therefore, $\left( x,y,z \right)=\left( 2,-3,4 \right)$