Answer to Question #150833 in Linear Algebra for shanto

Question

Solve the system using Gaussian elimination with back-substitution.

3x1 + 2x2 -x3= -15
5x1+ 3x2 + 2x3 =0
3x1+ x2 +3x3=11
11x1+7x2 = -30

Answer 1

3x1 + 2x2 -x3= -15

5x1+ 3x2 + 2x3 =0

3x1+ x2 +3x3=11

11x1+7x2 = -30

$\begin {bmatrix} 3 & 2 & -1 \ \ \ \ \ \ -15 \\ 5 & 3 & 2 \ \ \ \ \ \ \ \ \ \ \ \ \ 0\\ 3 & 1 & 3 \ \ \ \ \ \ \ \ \ \ \ \ 11 \\ 11 & 7 & 0 \ \ \ \ \ \ \ -30 \end{bmatrix} =$ $\begin {Vmatrix} R_2 =5R_1-3R_2 \\ R_3 = R_1-R_3 \\ R_4 = 5R_4-4R_2 \end{Vmatrix} =$ $\begin {bmatrix} 3 & 2 & -1 \ \ \ \ \ \ -15 \\ 0 & 1 & -1 \ \ \ \ \ \ \ -75\\ 0 & 1 & -2 \ \ \ \ \ \ \ \ \ \ \ 26 \\ 0 & 2 & -22 \ \ \ \ \ \ \ -150 \end{bmatrix}=$

$\begin {Vmatrix} R_1 =R_1-2R_2 \\ R_2 =R_2-R_3 \\ R_3 = 2R_3-R_4 \\ R_4 = 2R_2- R_4 \end{Vmatrix} =$ $\begin {bmatrix} 3 & 0 & 11 \ \ \ \ \ \ -135 \\ 0 & 1 & -1 \ \ \ \ \ \ \ -75\\ 0 & 0 & 18 \ \ \ \ \ \ \ \ \ \ \ 202 \\ 0 & 0 & 20 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{bmatrix}=$ $\begin {Vmatrix} R_1 =\frac{R_1}{3} \\ R_2 =R_2-R_3 \\ R_3 = \frac{R_3}{18} \\ R_4 = 10R_3-9R_4 \end{Vmatrix} =$

$\begin {bmatrix} 1 & 0 & \frac{11}{3} \ \ \ \ \ \ -\frac{135}{3} \\ 0 & 1 & -1 \ \ \ \ \ \ -75\\ 0 & 0 & 1 \ \ \ \ \ \ \ \ \ \ \frac{101}{9} \\ 0 & 0 & 0 \ \ \ \ \ \ \ \ \ 2020 \end{bmatrix};$

In $R_4$ $0 \ne 2020$ . So this system has no solution.