Answer
The solution of the system is $\frac{21}{13},\frac{10}{13},\frac{-2}{13}$
Work Step by Step
Given: $A=\begin{bmatrix}
5&-1 & 2\\
-2&6&9\\
-7& 5&-3
\end{bmatrix} $
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
5&-1 & 2|7 \\
-2&6&9 |0\\
-7& 5&-3 |-7
\end{bmatrix} \approx^1\begin{bmatrix}
1&11 & 20|7 \\
-2&6&9 |0\\
-7& 5&-3 |-7
\end{bmatrix} \approx^2\begin{bmatrix}
1&11 & 20|7 \\
0&28&49 |14\\
0& 82&137 |42
\end{bmatrix}\approx^3 \begin{bmatrix}
1&11 & 20|7 \\
0&1&\frac{7}{4} |\frac{1}{2}\\
0& 82&137 |42
\end{bmatrix} \approx^4 \begin{bmatrix}
1&11 & 20|7 \\
0&1&\frac{7}{4} |\frac{1}{2}\\
0& 0&-\frac{13}{2} |1
\end{bmatrix} \approx^5 \begin{bmatrix}
1&11 & 20|7 \\
0&1&\frac{7}{4} |\frac{1}{2}\\
0& 0&1 |-\frac{2}{13}
\end{bmatrix}$
$1.A_{21}(2)$
$2.A_{12}(2),A_{13}(7)$
$3. M_2(\frac{1}{8})$
$4.A_{23}(-82)$
$3. M_2(\frac{-2}{3})$
The equivalent system for this matrix is now:
$x_3=\frac{-2}{13}$
$x_2+\frac{7}{4}x_3 =\frac{1}{2} \rightarrow x_2=\frac{10}{13}$
$x_1-2x_2+x_3=3 \rightarrow x_1=\frac{21}{13}$
The solution of the system is $\frac{21}{13},\frac{10}{13},\frac{-2}{13}$
