Answer
The solution is $(-1,-2,-1,4)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
1&1 &1 &-4 |4\\
1 &-1 &-1 &-1| 2\\
1& 1 & -1 & 1 |-2\\
1 & -1 & 1 & 1|-8
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
1&1 &1 &-4 |4\\
1 &-1 &-1 &-1| 2\\
1& 1 & -1 & 1 |-2\\
1 & -1 & 1 & 1|-8
\end{bmatrix} \approx^1\begin{bmatrix}
1&1 &1 &-4 |4\\
0&-2 &-2 &0| -2\\
0& 0 & -2 & 2 |-6\\
0 & -2 & 0 & 2|-12
\end{bmatrix} \approx^2\begin{bmatrix}
1&1 &1 &-1 |4\\
0&1 &1 &0| 1\\
0& 0 &1 & -1 |3\\
0 & 1 & 0 & -1|6
\end{bmatrix} \approx^3\begin{bmatrix}
1&0 &0 &-1 |3\\
0&1 &1 &0| 1\\
0& 0 &1 & -1 |3\\
0 & 0 & -1 & -1|5
\end{bmatrix} \approx^4 \begin{bmatrix}
1&0 &0 &-1 |3\\
0&1 &0&1| -2\\
0& 0 &1 & -1 |3\\
0 & 0 & 0 & -2|8
\end{bmatrix} \approx^5 \begin{bmatrix}
1&0 &0 &-1 |3\\
0&1 &0&1| -2\\
0& 0 &1 & -1 |3\\
0 & 0 & 0 & 1|-4
\end{bmatrix} \approx^6 \begin{bmatrix}
1&0 &0 &0 |-1\\
0&1 &0&0| 2\\
0& 0 &1 & 0 |-1\\
0 & 0 & 0 & 1|-4
\end{bmatrix}$
$1.A_{12}(-1), A_{13}(-1),A_{14}(-1)$
$2.M_2(-\frac{1}{2}),M_3(-\frac{1}{2}),M_4(-\frac{1}{2})$
$3.A_{24}(-1),A_{32}(-1),A_{34}(1)$
$4.M_4(-\frac{1}{2})$
$5.A_{41}(1),A_{42}(-1),A_{43}(1)$
This matrix is now in row-echelon form. The equivalent system is:
$x_1=-1$
$x_2=2$
$x_3=-1$
$x_4=-4$
The solution is $(-1,-2,-1,4)$
