Answer
$(2,-1,3)$
Work Step by Step
Step 1. Based on the Cramer’s Rule, with the given equations, we can define the following determinants:
$\begin{array}( \\|D|= \\ \\ \end{array}
\begin{vmatrix}2 &-1&0\\5&0&3\\0&4&7 \end{vmatrix},
\begin{array}( \\|D_x|= \\ \\ \end{array}
\begin{vmatrix}5 &-1&0\\19&0&3\\17&4&7 \end{vmatrix},
\begin{array}( \\|D_y|= \\ \\ \end{array}
\begin{vmatrix}2 &5&0\\5&19&3\\0&17&7 \end{vmatrix},
\begin{array}( \\|D_z|= \\ \\ \end{array}
\begin{vmatrix}2 &-1&5\\5&0&19\\0&4&17 \end{vmatrix}$
Step 2. Evaluate the above determinants:
$|D|=(2)(0-12)-(5)(-7-0)=11$ (column1 expansion),
$|D_x|=(5)(0-12)-(-1)(133-51)=22$ (row1 expansion),
$|D_y|=(2)(133-51)-(5)(35-0)=-11$ (column1 expansion),
$|D_z|=(2)(0-76)-(5)(-17-20)=33$ (column1 expansion),
Step 3. Find the solutions as:
$x=\frac{|D_x|}{|D|}=2$,
$y=\frac{|D_y|}{|D|}=-1$,
$z=\frac{|D_z|}{|D|}=3$
