Answer
$(1,3,2)$
Work Step by Step
Step 1. Based on the Cramer’s Rule, with the given equations, we can define the following determinants:
$\begin{array}( \\|D|= \\ \\ \end{array}
\begin{vmatrix}2 &3&-5\\1&1&-1\\0&2&1 \end{vmatrix},
\begin{array}( \\|D_x|= \\ \\ \end{array}
\begin{vmatrix}1 &3&-5\\2&1&-1\\8&2&1 \end{vmatrix},
\begin{array}( \\|D_y|= \\ \\ \end{array}
\begin{vmatrix}2 &1&-5\\1&2&-1\\0&8&1 \end{vmatrix},
\begin{array}( \\|D_z|= \\ \\ \end{array}
\begin{vmatrix}2 &3&1\\1&1&2\\0&2&8 \end{vmatrix}$
Step 2. Evaluate the above determinants:
$|D|=(2)(1+2)-(1)(3+10)=-7$ (column1 expansion),
$|D_x|=(1)(1+2)-(2)(3+10)+(8)(-3+5)=-7$ (column1 expansion),
$|D_y|=(2)(2+8)-(1)(1+40)=-21$ (column1 expansion),
$|D_z|=(2)(8-4)-(1)(24-2)=-14$ (column1 expansion),
Step 3. Find the solutions as:
$x=\frac{|D_x|}{|D|}=1$,
$y=\frac{|D_y|}{|D|}=3$,
$z=\frac{|D_z|}{|D|}=2$
