Answer to Question #120102 in Algebra for Vikrant

Question

Solve the following system of linear equations
x+ y- 3z+ 2w = 0
2x -y +2z- 3w =0
3x- 2y+ z -4w =0
-4x +y -3z +w =0

Answer 1

$$A=\begin{pmatrix} 1 & 1 & -3 & 2 \\ 2 & -1 & 2 & -3 \\ 3 & -2 & 1 & -4 \\ -4 & 1 & -3 & 1 \end{pmatrix}, X=\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}, B=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

$$AX=B$$

$R_2=R_2-(2)R_1$

$$\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 & -3 & 8 & -7 \\ 3 & -2 & 1 & -4 \\ -4 & 1 & -3 & 1 \end{pmatrix}$$

$R_3=R_3-(3)R_1$

$$\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 & -3 & 8 & -7 \\ 0 & -5 & 10 & -10 \\ -4 & 1 & -3 & 1 \end{pmatrix}$$

$R_4=R_4+(4)R_1$

$$\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 & -3 & 8 & -7 \\ 0 & -5 & 10 & -10 \\ 0 & 5 & -15 & 9 \end{pmatrix}$$

$R_2=R_2/(-3)$

$$\begin{pmatrix} 1 & 1 & -3 & 2 \\ 0 &1 & -8/3 & 7/3 \\ 0 & -5 & 10 & -10 \\ 0 & 5 & -15 & 9 \end{pmatrix}$$

$R_1=R_1-R_2$

$$\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & -5 & 10 & -10 \\ 0 & 5 & -15 & 9 \end{pmatrix}$$

$R_3=R_3+(5)R_2$

$$\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & -10/3 & 5/3 \\ 0 & 5 & -15 & 9 \end{pmatrix}$$

$R_4=R_4-(5)R_2$

$$\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & -10/3 & 5/3 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}$$

$R_3=(-3/10)R_3$

$$\begin{pmatrix} 1 & 0 & -1/3 & -1/3 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}$$

$R_1=R_1+(1/3)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 1 & -8/3 & 7/3 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}$$

$R_2=R_2+(8/3)R_3$

$$\begin{pmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & -5/3 & -8/3 \end{pmatrix}$$

$R_4=R_4+(5/3)R_3$

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

$R_4=(-2/7)R_4$

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

$R_1=R_1+(1/2)R_4$

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

$R_2=R_2-R_4$

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

$R_3+R_3+(1/2)R_4$

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

$x=0$

$y=0$

$z=0$

$w=0$

$(0, 0,0,0)$