Question

Rework Problem 6.2 with the value of8changed to 4 .

Short answer

Therefore, the solution is$x_1=0.599,x_2=0.2673,and{x}_3=x_3=-0.0693$

Step-by-step solution

Given data.

The given matrix is,

$\left[\begin{array}{ccc}8& 2& 1\\ 4& 6& 2\\ 3& 4& 14\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right].......\left(1\right)$

Now, 8in first row is changed to 4 , therefore,

$\left[\begin{array}{ccc}4& 2& 1\\ 4& 6& 2\\ 3& 4& 14\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right].......\left(2\right)$

Determine the formulas of linear equation.

Write the formula of general form of linear equation.

$\mathbf{Ax}\mathbf{=}\mathbf{B}$ ……. (3)

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$N\times 1$matrix.

Here, N is order of the system.

Apply the Gauss Elimination method.

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

From matrix (1), the order of the matrix is 3, therefore,

$$\begin{gathered} N-1=3+1 \\ =2 \end{gathered}$$

Step 1: Now, in equation (1), divide first row by 1 and subtract from row 2 and

multiply first row by$\frac{3}{4}$and subtract from row 3 .

$\left[\begin{array}{ccc}4& 2& 1\\ 0& 4& 1\\ 0& 2.5& 13.25\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 0.25\end{array}\right]......\left(4\right)$

Step 2: Now, in equation (4), multiply second row by $\frac{5}{8}$and subtract from row 3 .

$\left[\begin{array}{ccc}4& 2& 1\\ 0& 4& 1\\ 0& 0& 12.625\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ -0.875\end{array}\right]......\left(5\right)$

Solve linear equations.

Now, from equation (3),

$$\begin{gathered} 12.625{\mathrm{x}}_3=-0.875......\left(6\right) \\ 4{\mathrm{x}}_2+{\mathrm{x}}_3=1......\left(7\right) \\ 4{\mathrm{x}}_1+2{\mathrm{x}}_2+{\mathrm{x}}_3=3......\left(8\right) \end{gathered}$$

Solve equation (6).

$$\begin{gathered} x_3=\frac{-0.875}{12.625} \\ =-0.0693 \end{gathered}$$

Substitute$-0.0693$for$x_3$in equation (7).

$$\begin{gathered} 4x_2+\left(-0.0+693\right)=1 \\ {x}_2=0.2673 \end{gathered}$$

Substitute$0.2673$for$x_2$and$-0.0693$for$x_3$in equation (8).

$$\begin{gathered} 4x_1+2\left(0.2673\right)+\left(-0.0693\right)=3 \\ {x}_1=0.599 \end{gathered}$$

Therefore, the solution is$x_1=0.0599,x_2=0.2673,and{x}_3=-0.0693$