Calculate: {\text{begin}array l x+3y-z=0 } 2x+y+z=1 3x-y+z=3\text{end}array .

Expression: $\left\{\begin{array} { l } x+3y-z=0 \\ 2x+y+z=1 \\ 3x-y+z=3\end{array} \right.$

Solve the equation for $x$

$\left\{\begin{array} { l } x=-3y+z \\ 2x+y+z=1 \\ 3x-y+z=3\end{array} \right.$

Substitute the given value of $x$ into the equation $2x+y+z=1$

$\left\{\begin{array} { l } 2\left( -3y+z \right)+y+z=1 \\ 3x-y+z=3\end{array} \right.$

Substitute the given value of $x$ into the equation $3x-y+z=3$

$\left\{\begin{array} { l } 2\left( -3y+z \right)+y+z=1 \\ 3\left( -3y+z \right)-y+z=3\end{array} \right.$

Simplify the expression

$\left\{\begin{array} { l } -5y+3z=1 \\ 3\left( -3y+z \right)-y+z=3\end{array} \right.$

Simplify the expression

$\left\{\begin{array} { l } -5y+3z=1 \\ -10y+4z=3\end{array} \right.$

Multiply both sides of the equation by $-2$

$\left\{\begin{array} { l } 10y-6z=-2 \\ -10y+4z=3\end{array} \right.$

Sum the equations vertically to eliminate at least one variable

$-2z=1$

Divide both sides of the equation by $-2$

$z=-\frac{ 1 }{ 2 }$

Substitute the given value of $z$ into the equation $-5y+3z=1$

$-5y+3 \times \left( -\frac{ 1 }{ 2 } \right)=1$

Solve the equation for $y$

$y=-\frac{ 1 }{ 2 }$

Substitute the given values of $\begin{array} { l }y,& z\end{array}$ into the equation $x=-3y+z$

$x=-3 \times \left( -\frac{ 1 }{ 2 } \right)+\left( -\frac{ 1 }{ 2 } \right)$

Simplify the expression

$x=1$

The possible solution of the system is the ordered triple $\left( x, y, z\right)$

$\left( x, y, z\right)=\left( 1, -\frac{ 1 }{ 2 }, -\frac{ 1 }{ 2 }\right)$

Check if the given ordered triple is a solution of the system of equations

$\left\{\begin{array} { l } 1+3 \times \left( -\frac{ 1 }{ 2 } \right)-\left( -\frac{ 1 }{ 2 } \right)=0 \\ 2 \times 1+\left( -\frac{ 1 }{ 2 } \right)+\left( -\frac{ 1 }{ 2 } \right)=1 \\ 3 \times 1-\left( -\frac{ 1 }{ 2 } \right)+\left( -\frac{ 1 }{ 2 } \right)=3\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } 0=0 \\ 1=1 \\ 3=3\end{array} \right.$

Since all of the equalities are true, the ordered triple is the solution of the system

$\left( x, y, z\right)=\left( 1, -\frac{ 1 }{ 2 }, -\frac{ 1 }{ 2 }\right)$