Answer
$x=\left[\begin{array}{c}3 \\ 4 \\ -6\end{array}\right]$
Work Step by Step
We substite $U x=y$ and first solve $L y=b$ using augmented matrix:
$\left[\begin{array}{ccc|c}1 & 0 & 0 & -7 \\ -1 & 1 & 0 & 5 \\ 2 & -5 & 1 & 2\end{array}\right]$
$y_{1}=-7$
$-y_{1}+y_{2}=5$
$7+y_{2}=5$
$y_{2}=-2$
$2 y_{1}-5 y_{2}+y_{3}=2$
$-14+10+y_{3}=2$
$y_{3}=6$
Now get $U x=y$ using augmented matrix:
$\left[\begin{array}{ccc|c}3 & -7 & -2 & -7 \\ 0 & -2 & -1 & -2 \\ 0 & 0 & -1 & 6\end{array}\right]$
$x_{3}=-6$
$-2 x_{2}-x_{3}=-2$
$-2 x_{2}+6=-2$
$-2 x_{2}=-8$
$x_{2}=4$
$3 x_{1}-7 x_{2}-2 x_{3}=-7$
$3 x_{1}-28+12=-7$
$3 x_{1}-16=-7$
$x_{1}=3$
Solve using augmented matrix by ordinary row reduction:
$\left[\begin{array}{ccc|c}3 & -7 & -2 & -7 \\ -3 & 5 & 1 & 5 \\ 6 & -4 & 0 & 2\end{array}\right],$ multiply first row with 1,-2 and add to second and
$\sim\left[\begin{array}{ccc|c}3 & -7 & -2 & -7 \\ 0 & -2 & -1 & -2 \\ 0 & 10 & 4 & 16\end{array}\right],$ multiply second row with 5 and add to third
$\sim\left[\begin{array}{ccc|c}3 & -7 & -2 & -7 \\ 0 & -2 & -1 & -2 \\ 0 & 0 & -1 & 6\end{array}\right]$
$x_{3}=-6$
$-2 x_{2}-x_{3}=-2$
$-2 x_{2}+6=-2$
$-2 x_{2}=-8$
$x_{2}=4$
$3 x_{1}-7 x_{2}-2 x_{3}=-7$
$3 x_{1}-28+12=-7$
$x_{1}=3$
Result is same, $x=\left[\begin{array}{c}3 \\ 4 \\ -6\end{array}\right]$
