Answer
The solution $(0,0,0)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
1&-1 &1 |0\\
0&3 &2 |0\\
3&0 &-1 | 0 \\
5 &1 &-1|0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
1&-1 &1 |0\\
0&3 &2 |0\\
3&0 &-1 | 0 \\
5 &1 &-1|0
\end{bmatrix} \approx^1
\begin{bmatrix}
1&-1 &1 |0\\
0&3 &2 |0\\
0&3 &-4 | 0 \\
0 &6 &-6|0
\end{bmatrix} \approx^2
\begin{bmatrix}
1&-1 &1 |0\\
0&1 &\frac{2}{3} |0\\
0&3 &-4 | 0 \\
0 &6 &-6|0
\end{bmatrix}
\approx^3
\begin{bmatrix}
1&0 &\frac{5}{3} |0\\
0&1 &\frac{2}{3} |0\\
0&0 &-6 | 0 \\
0 &0 &-10|0
\end{bmatrix}
\approx^4 \begin{bmatrix}
1&0 &\frac{5}{3} |0\\
0&1 &\frac{2}{3} |0\\
0&0 &1| 0 \\
0 &0 &-10|0
\end{bmatrix} \approx^5\begin{bmatrix}
1&0 &0 |0\\
0&1 &0|0\\
0&0 &1| 0 \\
0 &0 &0|0
\end{bmatrix}$
$1.A_{13}(-3),A_{14}(-5)$
$2.M_2(\frac{1}{3})$
$3.A_{21}(1),A_{23}(-3),A_{24}(-6)$
$4.M_3(-\frac{1}{6})$
$5.A_{31}(-\frac{5}{3}),A_{32}(-\frac{2}{3}),A_{34}(10)$
This matrix is now in row-echelon form.
The solution set is:
$x_1=0$
$x_2=0$
$x_3=0$
Hence, the solution is $(0,0,0)$
