## Precalculus (6th Edition) Blitzer

$\left( 2,3,5 \right)$
According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$. $D$ is the determinant and all three denominators consist of x, y and z co-efficients. $D=$ $\left| \begin{matrix} 1 & -3 & 1 \\ 1 & 2 & 0 \\ 2 & -1 & 0 \\ \end{matrix} \right|$ ${{D}_{x}}$ is the determinant in the numerator for x obtained by replacing the x- co-efficient in D with the constants on the right sides of the equations. ${{D}_{x}}=$ $\left| \begin{matrix} -2 & -3 & 1 \\ 8 & 2 & 0 \\ 1 & -1 & 0 \\ \end{matrix} \right|$ ${{D}_{y}}$ is the determinant in the numerator for y obtained by replacing the y- co-efficient in D with the constants on the right sides of the equations. ${{D}_{y}}=$ $\left| \begin{matrix} 1 & -2 & 1 \\ 1 & 8 & 0 \\ 2 & 1 & 0 \\ \end{matrix} \right|$ ${{D}_{z}}$ is the determinant in the numerator for z obtained by replacing the z- co-efficient in D with the constants on the right sides of the equations. ${{D}_{z}}=$ $\left| \begin{matrix} 1 & -3 & -2 \\ 1 & 2 & 8 \\ 2 & -1 & 1 \\ \end{matrix} \right|$. Calculate the four determinants. \begin{align} & D=\left| \begin{matrix} 1 & -3 & 1 \\ 1 & 2 & 0 \\ 2 & -1 & 0 \\ \end{matrix} \right| \\ & =1\left( -1-4 \right) \\ & =-5 \end{align} \begin{align} & {{D}_{x}}=\left| \begin{matrix} -2 & -3 & 1 \\ 8 & 2 & 0 \\ 1 & -1 & 0 \\ \end{matrix} \right| \\ & =1\left( -8-2 \right) \\ & =-10 \end{align} \begin{align} & {{D}_{y}}=\left| \begin{matrix} 1 & -2 & 1 \\ 1 & 8 & 0 \\ 2 & 1 & 0 \\ \end{matrix} \right| \\ & =1\left( 1-16 \right) \\ & =-15 \end{align} \begin{align} & {{D}_{z}}=\left| \begin{matrix} 1 & -3 & -2 \\ 1 & 2 & 8 \\ 2 & -1 & 1 \\ \end{matrix} \right| \\ & =1\left( 2+8 \right)+3\left( 1-16 \right)-2\left( -1-4 \right) \\ & =-25 \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{-25}{-5} \\ & =5 \end{align} Therefore, $\left( x,y,z \right)=\left( 2,3,5 \right)$