Question

Using Gauss elimination, solve the following linear algebraic

equations:

$$\begin{gathered} \mathbf{-}\mathbf{25}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{5}{\mathbf{x}}_{\mathbf{2}}\mathbf{10}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}\mathbf{10}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{5}{\mathbf{x}}_{\mathbf{1}}\mathbf{-}\mathbf{10}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{5}{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{2} \\ \mathbf{10}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{5}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{10}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}\mathbf{10}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{1} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{10}}_{{\mathbf{x}}_{\mathbf{1}}}\mathbf{-}\mathbf{20}{\mathbf{x}}_{\mathbf{4}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Short answer

Therefore, the solution is $x_1=\frac{1}{7},x_2=-\frac{1}{35},x_3=\frac{1}{5}\mathrm{and}{x}_4=\frac{6}{35}$.

Step-by-step solution

Given data.

The given linear equations are,

$$\begin{gathered} -25{\mathrm{x}}_1+5{\mathrm{x}}_{2}10{\mathrm{x}}_3+10\mathrm{x}......\left(1\right) \\ 5{\mathrm{x}}_1-10{\mathrm{x}}_2+5{\mathrm{x}}_3=2.......\left(2\right) \\ 10{\mathrm{x}}_1+5{\mathrm{x}}_2-10{\mathrm{x}}_3+10{\mathrm{x}}_4=1.......\left(3\right) \\ {10}_{{\mathrm{x}}_1}-20{\mathrm{x}}_4=-2.......\left(4\right) \end{gathered}$$

Determine the formulas of linear equation.

Write the formula of general form of linear equation.

${\mathbf{A}}_{\mathbf{X}}\mathbf{=}\mathbf{B}$ ……. (1)

Here, $x$is the $N\times 1$matrix which is the required solution, A is a$N\times N$matrix

which represents the coefficients of x , and B is the $N\times N$matrix.

Here, N is order of the system.

Apply the Gauss Elimination method.

Write the equations (1), (2), (3) and (4) in matrix form.

$\left[\begin{array}{cccc}-25& 5& 10& 10\\ 5& -10& 5& 0\\ 10& 5& -10& 10\\ 10& 0& 0& -20\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 1\\ -2\end{array}\right]......\left(5\right)$

For Gauss Siedel elimination method, $N-1$steps are needed.

From matrix (5), the order of the matrix is 4, therefore,

$$\begin{gathered} N-1=4-1 \\ =3 \end{gathered}$$

Step 1: Now, in equation (5), divide first row by 5 and add them to rows 2,

multiply first row by 0.4 and add them to rows 3 and 4 .

$\left[\begin{array}{cccc}-25& 5& 10& 10\\ 0& -9& 7& 2\\ 0& 7& -6& 14\\ 0& 2& 4& -16\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 1\\ -2\end{array}\right]......\left(6\right)$

Step 2: Now, in equation (6), multiply second row by $\frac{7}{9}$and add to row 3 and

multiply second row by $\frac{2}{9}$and add to row 4.

$\left[\begin{array}{cccc}-25& 5& 10& 10\\ 0& -9& 7& 2\\ 0& 0& -\frac{5}{9}& \frac{140}{9}\\ 0& 0& \frac{50}{9}& \frac{140}{9}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ \frac{23}{9}\\ -\frac{14}{9}\end{array}\right]......\left(7\right)$

Step 3: Now, in equation (7), multiply third row by 10 and add to row 4 .

$\left[\begin{array}{cccc}-25& 5& 10& 10\\ 0& -9& 7& 2\\ 0& 0& -\frac{5}{9}& \frac{140}{9}\\ 0& 0& 0& \frac{140}{9}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ \frac{23}{9}\\ 24\end{array}\right]......\left(8\right)$

Solve linear equations.

Now, from equation (8),

$$\begin{gathered} 140x_4=24......\left(9\right) \\ -\frac{5}{9}{x}_3+\frac{140}{9}{x}_4=\frac{23}{9}.......\left(10\right) \\ -9{\mathrm{x}}_2+7{\mathrm{x}}_3+2{\mathrm{x}}_4=2.......\left(11\right) \\ -25{\mathrm{x}}_1+5{\mathrm{x}}_{2}10{\mathrm{x}}_3+10{\mathrm{x}}_4=0.......\left(12\right) \end{gathered}$$

Solve equation (9)

$$\begin{gathered} x_4=\frac{24}{140} \\ =\frac{6}{135} \end{gathered}$$

Substitute$\frac{6}{35}$for${\mathrm{x}}_4$in equation (10).

$$\begin{gathered} -\frac{5}{9}{x}_3\frac{140}{9}\left(\frac{6}{35}\right)=\frac{23}{9} \\ {x}_3=\frac{1}{5} \end{gathered}$$

Substitute$\frac{6}{35}$for$x_4$and$\frac{1}{5}$for$x_3$in equation (11).

$$\begin{gathered} -9x_2+7\left(\frac{1}{5}\right)+2\left(\frac{6}{35}\right)=2 \\ {x}_2=-\frac{1}{35} \end{gathered}$$

Substitute$\frac{6}{35}$for $x_{4,}-\frac{1}{35}$for$x_2$and$\frac{1}{5}$for$x_3$in equation (12).

$$\begin{gathered} -25x_1+5\left(-\frac{1}{35}\right)+10\left(\frac{1}{5}\right)+10\left(\frac{6}{35}\right)=0 \\ {x}_1=\frac{1}{7} \end{gathered}$$

Therefore, the solution is${\mathrm{x}}_1=\frac{1}{7},{\mathrm{x}}_2=-\frac{1}{35},{\mathrm{x}}_3=\frac{1}{5}\mathrm{and}{\mathrm{x}}_4=\frac{6}{35}$ .