Answer
The solution set is $(1,-3,-2)$
Work Step by Step
$Ax=b$
$\begin{bmatrix}
1 & -3 & 1\\
5 & -4 & 1\\
2 & 4 &-3
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}=\begin{bmatrix}
8\\
15\\
-4
\end{bmatrix}$
Converting the given system of equations to an augmented matrix and then using Gauss-Jordan elimination to determine the solution set to the given system.
$\begin{bmatrix}
1 & -3 & 1|8\\
5 & -4 & 1|15\\
2 & 4 &-3|-4
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & -3 & 1|8\\
0& 11 &-4|15\\
0 & 10 &-5|-4
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & -3 & 1|8\\
0& 1 &1|-5\\
0 & 10 &-5|-4
\end{bmatrix} \approx^3 \begin{bmatrix}
1 & 0 & 4|-7\\
0& 1 &1|-5\\
0 & 0 &-15|30
\end{bmatrix} \approx^4 \begin{bmatrix}
1 & 0 & 4|-7\\
0& 1 &1|-5\\
0 & 0 &1|-2
\end{bmatrix} \approx^5 \begin{bmatrix}
1 & 0 & 0|1\\
0& 1 &0|-3\\
0 & 0 &1|-2
\end{bmatrix}$
1. $A_{12}(-5),A_{13}(-2)$
2. $A_{32}(-1)$
3. $A_{21}(3), A_{23}(-10)$
4. $M_3(\frac{-1}{15})$
5. $A_{31}(-4),A_{32}(-1)$
The solution set is $(1,-3,-2)$
