Answer
$\{(1,-2,3)\}$
Work Step by Step
Use Cramer's rule, we have:
1. $D=\begin{vmatrix} 1& 1&-1 \\ 2 & -3 &-1 \\ 1 & 2 & 2 \end{vmatrix}=1((-3)(2)-(2)(-1))-1((2)(2)-(1)(-1))+(-1)((2)(2)-(1)(-3))=-16$
2. $D_x=\begin{vmatrix} -4& 1&-1 \\ 5 & -3 &-1 \\ 3 & 2 & 2 \end{vmatrix}=(-4)((-3)(2)-(2)(-1))-1((5)(2)-(3)(-1))+(-1)((5)(2)-(3)(-3))=-16$
3. $D_y=\begin{vmatrix} 1& -4&-1 \\ 2 & 5 &-1 \\ 1 & 3 & 2 \end{vmatrix}=1((5)(2)-(3)(-1))-(-4)((2)(2)-(1)(-1))+(-1)((2)(3)-(1)(5))=32$
4. $D_z=\begin{vmatrix} 1& 1&-4 \\ 2 & -3 &5 \\ 1 & 2 & 3 \end{vmatrix}=1((-3)(3)-(2)(5))-1((2)(3)-(1)(5))+(-4)((2)(2)-(1)(-3))=-48$
5. $x=\frac{D_x}{D}=\frac{-16}{-16}=1, y=\frac{D_y}{D}=\frac{32}{-16}=-2$ and $z=\frac{D_z}{D}=\frac{-48}{-15}=3$
6. Thus the solution set is $\{(1,-2,3)\}$
