Answer
The solution is, ${{x}_{1}}=2-37t,{{x}_{2}}=16t,{{x}_{3}}=1-7t,{{x}_{4}}=t $.
Work Step by Step
Consider the given system of equations
$\begin{align}
& {{x}_{1}}+4{{x}_{2}}+3{{x}_{3}}-6{{x}_{4}}=5 \\
& {{x}_{1}}+3{{x}_{2}}+{{x}_{3}}-4{{x}_{4}}=3 \\
& 2{{x}_{1}}+8{{x}_{2}}+7{{x}_{3}}-5{{x}_{4}}=11 \\
& 2{{x}_{1}}+5{{x}_{2}}-6{{x}_{4}}=4
\end{align}$
Therefore, in matrix form the system of equations can be written as
$ AX=b $
Where
$ A=\left[ \begin{array}{*{35}{r}}
1 & 4 & 3 & -6 \\
1 & 3 & 1 & -4 \\
2 & 8 & 7 & -5 \\
2 & 5 & 0 & -6 \\
\end{array} \right];b=\left[ \begin{array}{*{35}{r}}
5 \\
3 \\
11 \\
4 \\
\end{array} \right];X=\left[ \begin{matrix}
{{x}_{1}} \\
{{x}_{2}} \\
{{x}_{3}} \\
{{x}_{4}} \\
\end{matrix} \right]$
Consider the augmented matrix
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & 4 & 3 & -6 & 5 \\
1 & 3 & 1 & -4 & 3 \\
2 & 8 & 7 & -5 & 11 \\
2 & 5 & 0 & -6 & 4 \\
\end{array} \right]$
By applying elementary row operation on $ A $ we will convert it to its equivalent upper triangular matrix form.
Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-2{{R}_{1}},{{{R}'}_{4}}={{R}_{4}}-2{{R}_{1}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & 4 & 3 & -6 & 5 \\
0 & -1 & -2 & 2 & -2 \\
0 & 0 & 1 & 7 & 1 \\
0 & -3 & -6 & 6 & -6 \\
\end{array} \right]$
Step 2: Apply the operation ${{{R}'}_{4}}={{R}_{4}}-3{{R}_{3}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
1 & 4 & 3 & -6 & 5 \\
0 & -1 & -2 & 2 & -2 \\
0 & 0 & 1 & 7 & 1 \\
0 & 0 & 0 & 0 & 0 \\
\end{array} \right]$
Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $ A $ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=3$
