Answer
The solution is $(0,0,0,0)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
1&1 &-1 & 1 |0\\
1&1 &1 &-1 |0\\
3 &-1 &1 & -2|0\\
4 & 2 &-1 &1 |0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2&1 &-1 & 1 |0\\
1&1 &1 &-1 |0\\
3 &-1 &1 & -2|0\\
4 & 2 &-1 &1 |0
\end{bmatrix} \approx^1
\begin{bmatrix}
1&1 &1 &-1|0\\
2&1 &-1 & 1 |0\\
3 &-1 &1 & -2|0\\
4 & 2 &-1 &1 |0
\end{bmatrix} \approx^2 \begin{bmatrix}
1&1 &1 &-1|0\\
0&-1 &-3 & 3 |0\\
0 &-4 &-2 & 1|0\\
0& -2 &-5 &5 |0
\end{bmatrix}
\approx^3\begin{bmatrix}
1&1 &1 &-1|0\\
0&1 &3 & -3 |0\\
0 &-4 &-2 & 1|0\\
0& -2 &-5 &5 |0
\end{bmatrix} \approx^4 \begin{bmatrix}
1&0 &-2&2|0\\
0&1 &3 & -3 |0\\
0 &0 &10 & -11|0\\
0& 0 &-3 &3 |0
\end{bmatrix} \approx^5 \begin{bmatrix}
1&0 &-2&2|0\\
0&1 &3 & -3 |0\\
0& 0 &-3 &3|0\\
0 &0 &10 & -11 |0
\end{bmatrix} \approx^6\begin{bmatrix}
1&0 &-2&2|0\\
0&1 &3 & -3 |0\\
0& 0 &1 &-1|0\\
0 &0 &10 & -11 |0
\end{bmatrix} \approx^7 \begin{bmatrix}
1&0 &0&0|0\\
0&1 &0 & 0|0\\
0& 0 &1 &-1|0\\
0 &0 &0& -1 |0
\end{bmatrix} \approx^8 \begin{bmatrix}
1&0 &0&0|0\\
0&1 &0 & 0|0\\
0& 0 &1 &-1|0\\
0 &0 &0& 1 |0
\end{bmatrix} \approx^9 \begin{bmatrix}
1&0 &0&0|0\\
0&1 &0 & 0|0\\
0& 0 &1 &0|0\\
0 &0 &0& 1 |0
\end{bmatrix}$
This matrix is now in row-echelon form.
The solution set is:
$x_1=0$
$x_2=0$
$x_3=0$
$x_4=0$
Hence, the solution is $(0,0,0,0)$
