Answer
$(4,2,-1)$
Work Step by Step
Step 1. Based on the Cramer’s Rule, with the given equations, we can define the following determinants:
$\begin{array}( \\|D|= \\ \\ \end{array}
\begin{vmatrix} 1 &-1&2\\3&0 &1\\-1&2 &0 \end{vmatrix},
\begin{array}( \\|D_x|= \\ \\ \end{array}
\begin{vmatrix} 0&-1&2\\11&0 &1\\0&2 &0 \end{vmatrix},
\begin{array}( \\|D_y|= \\ \\ \end{array}
\begin{vmatrix} 1 &0&2\\3&11 &1\\-1&0 &0 \end{vmatrix},
\begin{array}( \\|D_z|= \\ \\ \end{array}
\begin{vmatrix} 1 &-1&0\\3&0 &11\\-1&2 &0 \end{vmatrix}$
Step 2. Evaluate the above determinants:
$|D|=(1)(0-2)-(-1)(0+1)+(2)(6+0)=11$ (row1 expansion),
$|D_x|=-(11)((0-4)=44$ (column1 expansion),
$|D_y|=(-1)(0-22)=22$ (row3 expansion),
$|D_z|=(1)(0-22)-(-1)(0+11)=-11$ (row1 expansion),
Step 3. Find the solutions as:
$x=\frac{|D_x|}{|D|}=4$,
$y=\frac{|D_y|}{|D|}=2$,
$z=\frac{|D_z|}{|D|}=-1$
