Answer
The solution to the system of equations is $ x=-\frac{4}{7}-\frac{4}{7}t,y=\frac{5}{7}+\frac{5}{7}t,z=t $.
Work Step by Step
Consider the following system of equations:
$\begin{align}
& x-2y+2z=-2 \\
& 2x+3y-z=1
\end{align}$
Express the above system in the form of a matrix as below:
$\left[ \begin{matrix}
1 & -2 & 2 & -2 \\
2 & 3 & -1 & 1 \\
\end{matrix} \right]$
Using the elementary row operations we will obtain the echelon form of a matrix:
${{R}_{2}}\to {{R}_{2}}+\left( -2 \right){{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & -2 & 2 & -2 \\
0 & 7 & -5 & 5 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{1}{7}{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & -2 & 2 & -2 \\
0 & 1 & -\frac{5}{7} & \frac{5}{7} \\
\end{matrix} \right]$
Rewrite the system corresponding to echelon form of the matrix:
$ x-2y+2z=-2$ (I)
$ y-\frac{5}{7}z=\frac{5}{7}$ (II)
Since, the system has two equations and three variables, one variable can be chosen arbitrarily.
Let $ z=t $, then equation (II) gives,
$ y-\frac{5}{7}\left( t \right)=\frac{5}{7}$
Simplify,
$ y=\frac{5}{7}+\frac{5}{7}t $
Substitute the values of y and z in equation (I) to obtain:
$ x-2\left( \frac{5}{7}+\frac{5}{7}t \right)+2\left( t \right)=-2$
It can be further simplified as:
$ x=-\frac{4}{7}-\frac{4}{7}t $
Hence, the solution to the system is $ x=-\frac{4}{7}-\frac{4}{7}t,y=\frac{5}{7}+\frac{5}{7}t,z=t $.
