Answer to Question #147719 in Linear Algebra for Daher

Question

Solve the system of following equation by guass elimination method
3x1+2x2+6x3=2
6x1-7x2-11x3=8
-5x1+9x2+3x3=10

Answer 1

Let us solve the system of following equation by Guass elimination method:

$\begin{cases}3x_1+2x_2+6x_3=2\\ 6x_1-7x_2-11x_3=8\\ -5x_1+9x_2+3x_3=10 \end{cases}$

The augmented matrix  of the systems is the following:

$\left[ \begin{array}{cccc} 3 & 2 & 6 & 2\\ 6 & -7 & -11 & 8\\ -5 & 9 & 3 & 10 \end{array} \right] {r_2\to r_2-2r_1\choose r_3\to r_3+2r_1} \left[ \begin{array}{cccc} 3 & 2 & 6 & 2\\ 0 & -11 & -23 & 4\\ 1 & 13 & 15 & 14 \end{array} \right] {r_1\to r_1-3r_3}$

$\left[ \begin{array}{cccc} 0 & -37 & -39 & -40\\ 0 & -11 & -23 & 4\\ 1 & 13 & 15 & 14 \end{array} \right] {r_1\to r_1-3r_2} \left[ \begin{array}{cccc} 0 & -4 & 30 & -52\\ 0 & -11 & -23 & 4\\ 1 & 13 & 15 & 14 \end{array} \right] {r_2\to r_2-3r_1}$

$\left[ \begin{array}{cccc} 0 & -4 & 30 & -52\\ 0 & 1 & -113 & 160\\ 1 & 13 & 15 & 14 \end{array} \right] {r_1\to r_1+4r_2} \left[ \begin{array}{cccc} 0 & 0 & -422 & 588\\ 0 & 1 & -113 & 160\\ 1 & 13 & 15 & 14 \end{array} \right]$

Consequently, we have the following system:

$\begin{cases}-422x_3=588\\ x_2-113x_3=160\\ x_1+13x_2+15x_3=14 \end{cases}$

$\begin{cases}x_3=-\frac{294}{211}\\ x_2=160+113(-\frac{294}{211})\\ x_1=14-13x_2-15(-\frac{294}{211}) \end{cases}$

$\begin{cases}x_3=-\frac{294}{211}\\ x_2=\frac{538}{211}\\ x_1=14-13(\frac{538}{211})-15(-\frac{294}{211}) \end{cases}$

$\begin{cases}x_3=-\frac{294}{211}\\ x_2=\frac{538}{211}\\ x_1=\frac{370}{211} \end{cases}$