# Chapter 11 - Systems of Equations and Inequalities - Chapter Review - Review Exercises - Page 794: 31

$\{(-1,2,-3)\}$

#### Work Step by Step

We are given the system of equations: $\begin{cases} x+2y-z=6\\ 2x-y+3z=-13\\ 3x-2y+3z=-16 \end{cases}$ Compute $D,D_x,D_y,D_z$: $D=\begin{vmatrix}1&2&-1\\2&-1&3\\3&-2&3\end{vmatrix}=1(-3+6)-2(6-9)+(-1)(-4+3)=10$ $D_x=\begin{vmatrix}6&2&-1\\-13&-1&3\\-16&-2&3\end{vmatrix}=6(-3+6)-2(-39+48)+(-1)(26-16)=-10$ $D_y=\begin{vmatrix}1&6&-1\\2&-13&3\\3&-16&3\end{vmatrix}=1(-39+48)-6(6-9)+(-1)(-32+39)=20$ $D_z=\begin{vmatrix}1&2&6\\2&-1&-13\\3&-2&-16\end{vmatrix}=1(16-26)-2(-32+39)+6(-4+3)=-30$ Determine $x$: $x=\dfrac{D_x}{D}=\dfrac{-10}{10}=-1$ Determine $y$: $y=\dfrac{D_y}{D}=\dfrac{20}{10}=2$ Determine $z$: $z=\dfrac{D_z}{D}=\dfrac{-30}{10}=-3$ The solution set is: $\{(-1,2,-3)\}$

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