Answer
The solution is $(3-1,5)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2 &-1 &-1 |2\\
4 &3 &-2| -1\\
1& 4 & 1 |4
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2 &-1 &-1 |2\\
4 &3 &-2| -1\\
1& 4 & 1 |4
\end{bmatrix} \approx^1 \begin{bmatrix}
1& 4 & 1 |4\\
4 &3 &-2| -1\\
2 &-1 &-1 |2
\end{bmatrix} \approx^2 \begin{bmatrix}
1& 4 & 1 |4\\
0 &-13 &-6| -17\\
0&-9&-3 |-6
\end{bmatrix} \approx^3 \begin{bmatrix}
1& 4 & 1 |4\\
0&-9&-3 |-6\\
0 &-13 &-6| -17
\end{bmatrix} \approx^4 \begin{bmatrix}
1& 4 & 1 |4\\
0&1&\frac{1}{3} |\frac{2}{3}\\
0 &-13 &-6| -17
\end{bmatrix} \approx^5 \begin{bmatrix}
1& 4 & 1 |4\\
0&1&\frac{1}{3} |\frac{2}{3}\\
0 &0 &-\frac{5}{3}|- \frac{25}{3}
\end{bmatrix} \approx^6 \begin{bmatrix}
1& 4 & 0 |-1\\
0&1&0 |-1\\
0 &0 &-\frac{5}{3}|- \frac{25}{3}
\end{bmatrix} \approx^7 \begin{bmatrix}
1& 0 & 0 |3\\
0&1&0 |-1\\
0 &0 &1|5
\end{bmatrix}$
$1.P_{13}$
$2.A_{12}(-4), A_{13}(-2)$
$3.P_{23}$
$4.M_2(-\frac{1}{9})$
$5.A_{23}(\frac{1}{13})$
$6.A_{31}(\frac{3}{5}), A_{32}(\frac{1}{5})$
$7.A_{21}(-4),M_3(-\frac{3}{5})$
This matrix is now in row-echelon form. The equivalent system is:
$x_1=3$
$x_2=-1$
$x_3=5$
The solution is $(3-1,5)$
