We can use the Crout method for solving sots or linear simultaneous equations.
$a_{11}x_1 +a_{12}x_2+...+a_{1n}x_n=c_1$

$a_{21}x_1 +a_{22}x_2+...+a_{2n}x_n=c_2$

$\vdots \ldots \vdots \ldots \vdots$

$a_{n1}x_1 +a_{n2}x_2+...+a_{nn}x_n=c_n$

The augmented matrix is
$M=[A|C]= \left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ddots & \vdots \\ \vdots & \dots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array} \right| \left| \begin{array}{c} c_1 \\ c_2 \\ \vdots \\ c_n \end{array} \right]$

The steps for proceeding from
$M$
to
$M'$
are: elements of the first column, then elements of the first row to the right of the first column; elements of the second column, then elements of the second row to the right of the first column, and so on until all elements are determined. The first column of
$M'=[A'|C']$
is identical to the first column of
$M$
. Each element of the first row of
$M'$
except the first is obtained by dividing the corresponding element of
$M$
$a_{11}$
.
Each element
$a'_{ij}$
on or below the main diagonal of
$M'$
is obtained by subtracting from the corresponding element
$a_{ij}$
of
$M$
the sum of the products of elements in the ith row and corresponding elements in the jth column of
$M'$
, all uncalculated elements being imagined to be zero:
$a'_{OJ}=a_{ij}- \sum_{k=1}^{j-1}a'_{ik}a'_{kj}, \: j \leq i$
.
Each element
$a'_{ij}$
to the right of the principal diagonal is calculated by using the procedure above and then dividing by the diagonal element
$a'_{ii}$
of
$M'$
:
$a'_{ij}=\frac{a_{ij}- \sum a'_{ik}a'_{kj}}{a'_{ii}}, \: i \lt j$

The auxiliary matrix is
$M'= \left[ \begin{array}{cccc} a'_{11} & a'_{12} & \ldots & a'_{1n} \\ a'_{21} & a'_{22} & \ddots & \vdots \\ \vdots & \vdots & \ddots & \vdots \\ a'_{n1} & a'_{n2} & \ldots & a'_{nn} \end{array} \right| \left| \begin{array}{c} c'_1 \\ c'_2 \\ \vdots \\ c'_n \end{array} \right]$

The final solution vector
$\mathbf{x}$
is obtained from
$M'$
using the following three rules: 1. The elements of
$\mathbf{x}$
are determined in reverse order.
2. The last element
$x_n$
of
$\mathbf{x}$
is taken to be equal to
$c'_n$

3. Each remaining element
$x_i$
is obtained by subtracting from
$c'_i$
the sun of the products of elements in the ith row of
$A'$
and the corresponding element of
$\mathbf{x}$
. All uncalculated elements of
$\mathbf{x}$
are imagined to be zero:
$x_i =c'_i-\sum_{k=1}^n a'_{ik}a'_{kj}$

We can check the answer using the following:
1. In the auxiliary matrix, any element of the check column should exceed by unity the sum of the other elements in its row which lie to the right of the principle diagonal.
2. Each element of the check column associated with the solution vector should exceed by unity the corresponding value of the solution vector.
Example Solve the following problem using the Crout method.
$554.11x_1-281.91x_2-34.240x_3=273.02$

$-281.91x_1+226.81x_2-38.100x_3=-63.965$

$-34.240x_1+38.100x_2+80.221x_3=34.717$

The augmented matrix (the last column is a check column) is
$M= \left[ \begin{array}{ccc} 554.11 & -281.91 & -34.240 \\ -281.91 & 226.91 & 38.100 \\ -34.240 & 38.100 & 80.221 \end{array} \right| \left| \begin{array}{c} 273.02 \\ -63.965 \\ 34.717 \end{array} \right] \: \: \left( \begin{array}{c} 510.98 \\ -80.965 \\ 118.80 \end{array} \right)$

The auxiliary matrix and the check column are
$M'= \left[ \begin{array}{ccc} 554.11 & -0.50876 & -0.061793 \\ -281.91 & 83.385 & 0.24801 \\ -34.240 & 20.680 & 72.976 \end{array} \right| \left| \begin{array}{c} 0.49272 \\ 0.89870 \\ 0.45224 \end{array} \right] \: \: \left( \begin{array}{c} 0.92216 \\ 2.14668 \\ 1.45228 \end{array} \right)$

Then
$x_3=0.452244$

$x+2=0.8987-0.24801 \times 0.45224=-0.78654$

$x_1=0.49272-(-0.50876) \times 0.78654 - (-0.061793) \times 0.45224 =0.92083$