In Problems$29-31$, use Cramer’s Rule, if applicable, to solve each system.

role="math" localid="1646930838552" $\left\{\begin{array}{l}x+2y-z=6\\ 2x-y+3z=-13\\ 3x-2y+3z=-16\end{array}\right.$

The given system of equation is

$\left\{\begin{array}{l}x+2y-z=6\\ 2x-y+3z=-13\\ 3x-2y+3z=-16\end{array}\right.$

The determinant of coefficients of variables in a system of equations is

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 3& -2& 3\end{array}\right| \\ D=1\left|\begin{array}{cc}-1& 3\\ -2& 3\end{array}\right|-2\left|\begin{array}{cc}2& 3\\ 3& 3\end{array}\right|+(-1)\left|\begin{array}{cc}2& -1\\ 3& -2\end{array}\right| \\ D=1(-3+6)-2(6-9)-1(-4+3) \\ D=3+6+1 \\ D=10 \end{gathered}$$

Determinant$D\ne 0$, so according to Cramer's rule

$x=\frac{D_x}{D},y=\frac{D_y}{D},z=\frac{D_y}{D}$

Substitute constants of the right side of equal sign for the first column of D

role="math" localid="1646932690134" $$\begin{gathered} D_x=\left|\begin{array}{ccc}6& 2& -1\\ -13& -1& 3\\ -16& -2& 3\end{array}\right| \\ {D}_x=6\left|\begin{array}{cc}-1& 3\\ -2& 3\end{array}\right|-2\left|\begin{array}{cc}-13& 3\\ -16& 3\end{array}\right|+(-1)\left|\begin{array}{cc}-13& -1\\ -16& -2\end{array}\right| \\ {D}_x=6(-3+6)-2(-39+48)-1(26-16) \\ {D}_x=18-18-10 \\ {D}_x=-10 \end{gathered}$$

sorole="math" localid="1646933223722" $$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-10}{10} \\ x=-1 \end{gathered}$$

Substitute constants of the right side of equal sign for the second column of D

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& 6& -1\\ 2& -13& 3\\ 3& -16& 3\end{array}\right| \\ {D}_y=1\left|\begin{array}{cc}-13& 3\\ -16& 3\end{array}\right|-6\left|\begin{array}{cc}2& 3\\ 3& 3\end{array}\right|+(-1)\left|\begin{array}{cc}2& -13\\ 3& -16\end{array}\right| \\ {D}_y=1(-39+48)-6(6-9)-1(-32+39) \\ {D}_y=9+18-7 \\ {D}_y=20 \end{gathered}$$

so

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{20}{10} \\ y=2 \end{gathered}$$

Substitute constants of the right side of equal sign for the third column of D

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& 2& 6\\ 2& -1& -13\\ 3& -2& -16\end{array}\right| \\ {D}_z=1\left|\begin{array}{cc}-1& -13\\ -2& -16\end{array}\right|-2\left|\begin{array}{cc}2& -13\\ 3& -16\end{array}\right|+6\left|\begin{array}{cc}2& -1\\ 3& -2\end{array}\right| \\ {D}_z=1(16-26)-2(-32+39)+6(-4+3) \\ {D}_z=-10-14-6 \\ {D}_z=-30 \end{gathered}$$

So

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{-30}{10} \\ z=-3 \end{gathered}$$

Solution of system is$(-1,2,-3)$