Answer
$(x,y,z)=(13,0,-3)$.
Work Step by Step
The augmented matrix is
$\begin{bmatrix}
1&0 &5 & -2\\
0 & -2 & 1 & -3\\
0& 0 & -4 & 12
\end{bmatrix}$.
For the linear equation.
First column $=$ coefficients of $x$.
Second column $=$ coefficients of $y$
Third column $=$ coefficients of $z$
and fourth column $=$ the right-hand side.
The given matrix is a upper triangular form.
Rewrite the third row as a linear equation.
$\Rightarrow 0x+0y-4z=12$
$\Rightarrow -4z=12$
Divide both sides by $-4$.
$\Rightarrow \frac{-4z}{-4}=\frac{12}{-4}$
Simplify.
$\Rightarrow z=-3$
Rewrite the second row as a linear equation.
$\Rightarrow 0x-2y+1z=-3$
$\Rightarrow -2y+z=-3$
Substitute $z=-3$.
$\Rightarrow -2y+(-3)=-3$
Clear the parentheses.
$\Rightarrow -2y-3=-3$
Add $3$ to both sides.
$\Rightarrow -2y-3+2=-3+3$
Simplify.
$\Rightarrow -2y=0$
Divide both sides by $-2$.
$\Rightarrow \frac{-2y}{-2}=\frac{0}{-2}$
Simplify.
$\Rightarrow y=0$
Rewrite the first row as a linear equation.
$\Rightarrow 1x+0y+5z=-2$
$\Rightarrow x+5z=-2$
Substitute $z=-3$.
$\Rightarrow x+5(-3)=-2$
$\Rightarrow x-15=-2$
Add $15$ to both sides.
$\Rightarrow x-15+15=-2+15$
Simplify.
$\Rightarrow x=13$
The solution set is $(x,y,z)=(13,0,-3)$.
