Answer
Using Gaussian elimination to solve
$\left\{ \begin{align}
3x+2y+-2z=0 & \\
x-y+z=5 & \\
\end{align} \right.$
We obtain the matrix
$\left[ \left. \begin{align}
& \begin{matrix}
1 & -1 & 1 \\
\end{matrix} \\
& \begin{matrix}
0 & 1 & -1 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
5 \\
-3 \\
\end{matrix} \right]$
Translating this matrix back into equation form gives
$\left\{ \begin{matrix}
\underline{x-y+z=5} & \text{Equation}\ 1 \\
\underline{y-z=-3} & \text{Equation}\ 2 \\
\end{matrix} \right.$
Solving Equation 2 for $ y $ in terms of $ z $ results in $ y=z-3$. Substituting this expression for $ y $ in Equation 1 gives $ x=2$. The system’s solution set is $\left\{ 2,\ z-3,\ z \right\}$.
Work Step by Step
Convert the given linear equation into matrix form, $\left[ \left. \begin{align}
& \begin{matrix}
1 & -1 & 1 \\
\end{matrix} \\
& \begin{matrix}
0 & 1 & -1 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
5 \\
-3 \\
\end{matrix} \right]$
Write it back in the linear equation, $\begin{align}
x-y+z=5 & \\
y-z=-3 & \\
\end{align}$
Write these equations for x and y in terms of z
$ y=z-3$ and $ x=2$
Hence the solution set is $\left\{ \left( 2,\ z-3,\ z \right) \right\}$.
