Answer
Solution set: $\{ ( 2$ , $\displaystyle \frac{1}{2}t-\frac{1}{2}$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Swap equations (the coefficient of x is 1 in Eq.2).
Augmented matrix:
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & 1\\
3 & 2 & -1 & | & 5
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow-3R_{1}+R_{3}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & 1\\
0 & -4 & 2 & | & 2
\end{array}\right]\left\{\begin{array}{l}
.\\
\div(-4).
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & 1\\
0 & 1 & -1/2 & | & -1/2
\end{array}\right]$
The system is consistent and does not have a unique solution.
Let $z=t, t\in \mathbb{R}$.
Substituting into row 2,
$y-\displaystyle \frac{1}{2}t=-\frac{1}{2}$
$y=\displaystyle \frac{1}{2}t-\frac{1}{2}$
Substituting into row 1,
$x+2(\displaystyle \frac{1}{2}t-\frac{1}{2})-t=1$
$x+t-1-t=1$
$x=1+1$
$x=2$
Solution set: $\{ ( 2$ , $\displaystyle \frac{1}{2}t-\frac{1}{2}$ , $t )$ , $t\in \mathbb{R}\}$
