Answer
The solution is $(10,-1,4)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2& -1 &-4 |5\\
3&2 &-5| 8\\
5 & 6 & -6|20\\
1 &1 & -3 | -3
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2& -1 &-4 |5\\
3&2 &-5| 8\\
5 & 6 & -6|20\\
1 &1 & -3 | -3
\end{bmatrix} \approx^1\begin{bmatrix}
1 &1 & -3 | -3\\
3&2 &-5| 8\\
5 & 6 & -6|20\\
2& -1 &-4 |5
\end{bmatrix} \approx^2\begin{bmatrix}
1 &1 & -3 | -3\\
0&-1 &4| 17\\
0 & 1 & 9|35\\
0& -3 &2 |11
\end{bmatrix} \approx^3 \begin{bmatrix}
1 &1 & -3 | -3\\
0&1 &-4| -17\\
0 & 1 & 9|35\\
0& -3 &2 |11
\end{bmatrix} \approx^4 \begin{bmatrix}
1 &1 & -3 | -3\\
0&1 &-4| -17\\
0 & 0 &13|52\\
0& 0 &-10 |-40
\end{bmatrix} \approx^5\begin{bmatrix}
1 &1 & -3 | -3\\
0&1 &-4| -17\\
0 & 0 &1|4\\
0& 0 &-10 |-40
\end{bmatrix} \approx^6 \begin{bmatrix}
1 &1 & -3 | -3\\
0&1 &-4| -17\\
0 & 0 &1|4\\
0& 0 &0 |0
\end{bmatrix} $
The equivalent system is:
$x_1+x_2-3 x_3=-3$
$x_2-4x_3=-17$
$x_3 =4$
Applying back substitution yields:
$x_3 =4$
$x_2-4.4=-28 \rightarrow x_2=-1$
$x_1+(-1)-3.4=-3\rightarrow x_1=10$
The solution is $(10,-1,4)$
