Answer
The solution is $(2,-1,3)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2& -1 &3 |14\\
3& 1 &-2| -1\\
7 & 2 & -3|3\\
5 &-1 & -2 | 5
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2& -1 &3 |14\\
3& 1 &-2| -1\\
7 & 2 & -3|3\\
5 &-1 & -2 | 5
\end{bmatrix} \approx^1\begin{bmatrix}
3& 1 &-2 |-1\\
2& -1 &3| 14\\
7 & 2 & -3|3\\
5 &-1 & -2 | 5
\end{bmatrix} \approx^2\begin{bmatrix}
1& 2 &-5 |-15\\
2& -1 &3| -14\\
7 & 2 & -3|3\\
5 &-1 & -2 | 5
\end{bmatrix} \approx^3 \begin{bmatrix}
1& 2 &-5 |-15\\
0& -5 &13| 44\\
0 & -12 & 32|108\\
0 &-11 & 23 | 80
\end{bmatrix} \approx^4 \begin{bmatrix}
1& 2 &-5 |-15\\
0& -12 &32| 108\\
0 & -5 & 13|44\\
0 &-11 & 23 | 80
\end{bmatrix} \approx^5 \begin{bmatrix}
1& 2 &-5 |-15\\
0& -1 &9| 28\\
0 & -5 & 13|44\\
0 &-11 & 23 | 80
\end{bmatrix} \approx^6 \begin{bmatrix}
1& 2 &-5 |-15\\
0& 1 &-9| -28\\
0 & -5 & 13|44\\
0 &-11 & 23 | 80
\end{bmatrix} \approx^7 \begin{bmatrix}
1& 2 &-5 |-15\\
0& 1 &-9| -28\\
0 &0 & -32|-96\\
0 &0 & -76 | -228
\end{bmatrix} \approx^8 \begin{bmatrix}
1& 2 &-5 |-15\\
0& 1 &-9| -28\\
0 &0 & 32|96\\
0 &0 & -76 | -228
\end{bmatrix} \approx^9 \begin{bmatrix}
1& 2 &-5 |-15\\
0& 1 &-9| -28\\
0 &0 & 1|3\\
0 &0 & 0 | 0
\end{bmatrix}$
The equivalent system is:
$x_1-2x_2-5 x_3=-15$
$x_2-9x_3=-28$
$x_3 =3$
Applying back substitution yields:
$x_3 =3$
$x_2-9.3=-28 \rightarrow x_2=-1$
$x_1-2(-1)-5.3=-15 \rightarrow x_1=2$
The solution is $(2,-1,3)$
