Answer
The solution is, $ x=-2,y=-1,z=0$.
Work Step by Step
Consider the given system of equation, $\begin{align}
& x-2y+z=0 \\
& y-3z=-1 \\
& 2y+5z=-2
\end{align}$
Therefore, the system of equations can be written in matrix form as below:
$ AX=b $
Where
$ A=\left[ \begin{array}{*{35}{r}}
1 & -2 & 1 \\
0 & 1 & -3 \\
0 & 2 & 5 \\
\end{array} \right];b=\left[ \begin{array}{*{35}{r}}
0 \\
-1 \\
-2 \\
\end{array} \right];X=\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]$
Consider the augmented matrix
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & -2 & 1 & 0 \\
0 & 1 & -3 & -1 \\
0 & 2 & 5 & -2 \\
\end{array} \right]$
By applying elementary row operation on $ A $ we will convert it to its equivalent upper triangular matrix form.
Step 1: Apply the operation ${{{R}'}_{3}}={{R}_{3}}-2{{R}_{2}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & -2 & 1 & 0 \\
0 & 1 & -3 & -1 \\
0 & 0 & 11 & 0 \\
\end{array} \right]$
Therefore, the system of equations reduces to
$\begin{align}
& x-2y+z=0 \\
& y-3z=-1 \\
& 11z=0
\end{align}$
From $11z=0$ we get
$\begin{align}
& 11z=0 \\
& z=0
\end{align}$
From $ y-3z=-1$ we get
$\begin{align}
& y=3z-1 \\
& =-1
\end{align}$
From $ x-2y+z=0$ we get
$\begin{align}
& x=2y-z \\
& =-2-0 \\
& =-2
\end{align}$
The solution is, $ x=-2,y=-1,z=0$.
