Answer
Solution set: $\{ ( 2t-\displaystyle \frac{5}{4}$ , $\displaystyle \frac{13}{4}$ , $t )$ , $t\in \mathbb{R}\}$
Work Step by Step
Augmented matrix:
$\left[\begin{array}{lllll}
1 & 1 & -2 & | & 2\\
3 & -1 & -6 & | & -7
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow-3R_{1}+R_{2}
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -2 & | & 2\\
0 & -4 & 0 & | & -13
\end{array}\right]\left\{\begin{array}{l}
.\\
\div(-4)
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -2 & | & 2\\
0 & 1 & 0 & | & 13/4
\end{array}\right]$
The system is consistent and does not have a unique solution.
Let $z=t, t\in \mathbb{R}$.
Substituting into row 2,
$y+0=\displaystyle \frac{13}{4}$
$y=\displaystyle \frac{13}{4}$
Substituting into row 1,
$x+\displaystyle \frac{13}{4}-2t=2$
$x=2t+2-\displaystyle \frac{13}{4}$
$x=2t-\displaystyle \frac{5}{4}$
Solution set: $\{ ( 2t-\displaystyle \frac{5}{4}$ , $\displaystyle \frac{13}{4}$ , $t )$ , $t\in \mathbb{R}\}$
