Question

Use the Gauss-Jordan Elimination method to solve the system of linear equations.

xa + 3xb + xc = 4

2xa + 2xb + xc = -1

2xa + 3xb + xc = 3

Answer 1

$$\begin{matrix} x_a+3x_b+x_c=4 \\ \\ 2x_a+2x_b+x_c=-1 \\ \\ 2x_a+3x_b+x_c=3 \\ \end{matrix}$$

The augmented matrix is

$$A=\begin{pmatrix} 1 & 3 & 1 & & 4 \\ 2 & 2 & 1 & & -1 \\ 2 & 3 & 1 & & 3 \\ \end{pmatrix}$$

$R_2=R_2-2R_1$

$$\begin{pmatrix} 1 & 3 & 1 & & 4 \\ 0 & -4 & -1 & & -9 \\ 2 & 3 & 1 & & 3 \\ \end{pmatrix}$$

$R_3=R_3-2R_1$

$$\begin{pmatrix} 1 & 3 & 1 & & 4 \\ 0 & -4 & -1 & & -9 \\ 0 & -3 & -1 & & -5 \\ \end{pmatrix}$$

$R_2=R_2/(-4)$

$$\begin{pmatrix} 1 & 3 & 1 & & 4 \\ 0 & 1 & 1/4 & & 9/4 \\ 0 & -3 & -1 & & -5 \\ \end{pmatrix}$$

$R_1=R_1-3R_2$

$$\begin{pmatrix} 1 & 0 & 1/4 & & -11/4 \\ 0 & 1 & 1/4 & & 9/4 \\ 0 & -3 & -1 & & -5 \\ \end{pmatrix}$$

$R_3=R_3+3R_2$

$$\begin{pmatrix} 1 & 0 & 1/4 & & -11/4 \\ 0 & 1 & 1/4 & & 9/4 \\ 0 & 0 & -1/4 & & 7/4 \\ \end{pmatrix}$$

$R_3=-4R_3$

$$\begin{pmatrix} 1 & 0 & 1/4 & & -11/4 \\ 0 & 1 & 1/4 & & 9/4 \\ 0 & 0 & 1 & & -7\\ \end{pmatrix}$$

$R_1=R_1-R_3/4$

$$\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 1/4 & & 9/4 \\ 0 & 0 & 1 & & -7\\ \end{pmatrix}$$

$R_2=R_2-R_3/4$

$$\begin{pmatrix} 1 & 0 & 0 & & -1 \\ 0 & 1 & 0 & & 4 \\ 0 & 0 & 1 & & -7\\ \end{pmatrix}$$

Then $x_a=-1, x_b=4, x_c=-7.$

The solution is $(-1, 4, -7).$