Answer to Question #346039 in Linear Algebra for Beza

Question

Find the solution x1+2x2-3x3+x4=0

3x1-x2+5x-3x4=0

2x1+x2+x4=0

Answer 1

Write down the augmented matrix

$$A=\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 3 & -1 & 5 & -1 &| & 0 \\ 2 & 1 & 0 & 1 &| & 0 \\ \end{pmatrix}$$

$R_2=R_2-3R_1$

$$\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 0 & -7 & 14 & -4 &| & 0 \\ 2 & 1 & 0 & 1 &| & 0 \\ \end{pmatrix}$$

$R_3=R_3-2R_1$

$$\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 0 & -7 & 14 & -4 &| & 0 \\ 0 & -3 & 6 & -1 &| & 0 \\ \end{pmatrix}$$

$R_2=-R_2/7$

$$\begin{pmatrix} 1 & 2 & -3 & 1 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & -3 & 6 & -1 &| & 0 \\ \end{pmatrix}$$

$R_1=R_1-2R_2$

$$\begin{pmatrix} 1 & 0 & 1 & -1/7 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & -3 & 6 & -1 &| & 0 \\ \end{pmatrix}$$

$R_3=R_3+3R_2$

$$\begin{pmatrix} 1 & 0 & 1 & -1/7 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & 0 & 0 & 5/7 &| & 0 \\ \end{pmatrix}$$

$R_3=7R_3/5$

$$\begin{pmatrix} 1 & 0 & 1 & -1/7 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & 0 & 0 & 1 &| & 0 \\ \end{pmatrix}$$

$R_1=R_1+R_3/7$

$$\begin{pmatrix} 1 & 0 & 1 & 0 &| & 0 \\ 0 & 1 & -2 & 4/7 &| & 0 \\ 0 & 0 & 0 & 1 &| & 0 \\ \end{pmatrix}$$

$R_2=R_2-4R_3/7$

$$\begin{pmatrix} 1 & 0 & 1 & 0 &| & 0 \\ 0 & 1 & -2 & 0 &| & 0 \\ 0 & 0 & 0 & 1 &| & 0 \\ \end{pmatrix}$$

Take $x_3=t, t\in \R.$

Then $x_1=-t, x_2=2t, x_3=t, x_4=0, t\in \R$

$$\{-t, 2t, t, 0\}, t\in \R$$