Answer
The solution is $(0,0,0)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2&1 &-1 |0\\
3 &-1 &2 | 0\\
1& -1 & -1 |0\\
5&2&-2
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2&1 &-1 |0\\
3 &-1 &2 | 0\\
1& -1 & -1 |0\\
5&2&-2
\end{bmatrix}\approx^1
\begin{bmatrix}
1&-1 &-1 |0\\
2 &1 &-1 | 0\\
1& -1 & -1 |0\\
5&2&-2
\end{bmatrix} \approx^2
\begin{bmatrix}
1&-1 &-1 |0\\
0 &2 &5 | 0\\
0&3 & 1 |0\\
0&7&3
\end{bmatrix}
\approx^3
\begin{bmatrix}
1&-1 &-1 |0\\
0 &3 &1 | 0\\
0&2 & 5 |0\\
0&7&3
\end{bmatrix}
\approx^4 \begin{bmatrix}
1&-1 &-1 |0\\
0 &1 &-4 | 0\\
0&2 & 5 |0\\
0&7&3
\end{bmatrix} \approx^5 \begin{bmatrix}
1&0 &-5 |0\\
0 &1 &-4 | 0\\
0&0 & 13 |0\\
0&0&31
\end{bmatrix} \approx^6 \begin{bmatrix}
1&0 &-5 |0\\
0 &1 &-4 | 0\\
0&0 & 1 |0\\
0&0&31
\end{bmatrix} \approx^7 \begin{bmatrix}
1&0 &0 |0\\
0 &1 &0 | 0\\
0&0 & 1 |0\\
0&0&0
\end{bmatrix} $
$1.P_{13}$
$2.A_{12}(-3), A_{13}(-2),A_{14}(-5)$
$3.P_{13}$
$4.A_{32}(-1)$
$5.A_{21}(1),A_{23}(-2),A_{24}(-7)$
$6.M_3(\frac{1}{13})$
$5.A_{31}(5),A_{32}(4),A_{34}(-31)$
This matrix is now in row-echelon form.
The solution set is:
$x_1=0$
$x_2=0$
$x_3=0$
The solution is $(0,0,0)$
