Answer
The solution is, ${{x}_{1}}=2,{{x}_{2}}=-2,{{x}_{3}}=3,{{x}_{4}}=4$
Work Step by Step
Consider the given system of equation
$\begin{align}
& 3{{x}_{1}}+5{{x}_{2}}-8{{x}_{3}}+5{{x}_{4}}=-8 \\
& {{x}_{1}}+2{{x}_{2}}-3{{x}_{3}}+{{x}_{4}}=-7 \\
& 2{{x}_{1}}+3{{x}_{2}}-7{{x}_{3}}+3{{x}_{4}}=-11 \\
& 4{{x}_{1}}+8{{x}_{2}}-10{{x}_{3}}+7{{x}_{4}}=-10
\end{align}$
Therefore, the system of equations can be written in matrix form as below:
$ AX=b $
Where
$ A=\left[ \begin{matrix}
3 & 5 & -8 & 5 \\
1 & 2 & -3 & 1 \\
2 & 3 & -7 & 3 \\
4 & 8 & -10 & 7 \\
\end{matrix} \right];b=\left[ \begin{array}{*{35}{r}}
-8 \\
-7 \\
-11 \\
-10 \\
\end{array} \right];X=\left[ \begin{matrix}
{{x}_{1}} \\
{{x}_{2}} \\
{{x}_{3}} \\
{{x}_{4}} \\
\end{matrix} \right]$
Consider the augmented matrix
$\left[ A|b \right]=\left[ \begin{matrix}
3 & 5 & -8 & 5 & -8 \\
1 & 2 & -3 & 1 & -7 \\
2 & 3 & -7 & 3 & -11 \\
4 & 8 & -10 & 7 & -10 \\
\end{matrix} \right]$
By applying the elementary row operation on $ A $ we will convert it to its equivalent upper triangular matrix form.
Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-\frac{1}{3}{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-\frac{2}{3}{{R}_{1}},{{{R}'}_{4}}={{R}_{4}}-\frac{4}{3}{{R}_{1}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
3 & 5 & -8 & 5 & -8 \\
0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & -\frac{13}{3} \\
0 & -\frac{1}{3} & -\frac{5}{3} & -\frac{1}{3} & -\frac{17}{3} \\
0 & \frac{4}{3} & \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\
\end{array} \right]$
Step 2: Apply the operation ${{{R}'}_{3}}={{R}_{3}}+{{R}_{2}},{{{R}'}_{4}}={{R}_{4}}-4{{R}_{2}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
3 & 5 & -8 & 5 & -8 \\
0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & -\frac{13}{3} \\
0 & 0 & -2 & -1 & -10 \\
0 & 0 & 2 & 3 & 18 \\
\end{array} \right]$
Step 3: Apply the operation ${{{R}'}_{4}}={{R}_{4}}+{{R}_{3}}$
$\left[ A|b \right]=\left[ \begin{array}{*{35}{r}}
3 & 5 & -8 & 5 & -8 \\
0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & -\frac{13}{3} \\
0 & 0 & -2 & -1 & -10 \\
0 & 0 & 0 & 2 & 8 \\
\end{array} \right]$
Therefore, the system of equations reduces to
$\begin{align}
& 3{{x}_{1}}+5{{x}_{2}}-8{{x}_{3}}+5{{x}_{4}}=-8 \\
& \frac{1}{3}{{x}_{2}}-\frac{1}{3}{{x}_{3}}-\frac{2}{3}{{x}_{4}}=-\frac{13}{3} \\
& -2{{x}_{3}}-{{x}_{4}}=-10 \\
& 2{{x}_{4}}=8 \\
\end{align}$
From (IV) we get,
${{x}_{4}}=4$
From (III) we get,
$\begin{align}
& {{x}_{3}}=-\frac{1}{2}\left( {{x}_{4}}-10 \right) \\
& =-\frac{1}{2}\left( 4-10 \right) \\
& =3
\end{align}$
From (II) we get,
$\begin{align}
& {{x}_{2}}={{x}_{3}}+2{{x}_{4}}-13 \\
& =3+8-13 \\
& =-2
\end{align}$
From (I) we get, $\begin{align}
& {{x}_{1}}=\frac{1}{3}\left( -5{{x}_{2}}+8{{x}_{3}}-5{{x}_{4}}-8 \right) \\
& =\frac{1}{3}\left( 10+24-20-8 \right) \\
& =2
\end{align}$
The solution of the system is ${{x}_{1}}=2,{{x}_{2}}=-2,{{x}_{3}}=3,{{x}_{4}}=4$.
