Answer
$x=\begin{pmatrix}
-11\\
7
\end{pmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
1 & 2\\
2&3
\end{bmatrix} $
$b=\begin{bmatrix}
3\\
-1
\end{bmatrix} $
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
1 & 2\\
2&3
\end{bmatrix} \approx^1\begin{bmatrix}
1 & 2\\
0&-1
\end{bmatrix}=U$
$1.A_{12}(-2)$
Corresponding multiplier: $m_{21}=2$
Consequently, we get $L=\begin{bmatrix}
1&0\\
2& 1
\end{bmatrix}$
We now solve the two triangular systems $Ly = b$ and $Ux = y$. Using forward substitution on the first of these systems, we have:
$y_1=3$
$2y_1+y_2=-1 \rightarrow 2.3+y_2=-1 \rightarrow y_2=-7$
Solving $Ux = y$ via back substitution yields:
$x_1+2x_2=3$
$-x_2=-7$
then $x_2=7$
$x_1+2.7=3 \rightarrow x_1=-11$
The solution of the system is $x=\begin{pmatrix}
-11\\
7
\end{pmatrix}$
