Answer to Question #298576 in Linear Algebra for shahana

Question

solve the linear system by gaussian elimination

 x − y + 2z − w = −1,

2x + y − 2z − 2w = −2 ,

−x + 2y − 4z + w = 1 ,

3x − 3w = −3

Answer 1

Augmented matrix

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

$R_2=R_2-2R_1$

$$\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 3 & -6 & 0 & & 0\\ -1 & 2 & -4 & 1 & & 1 \\ 3 & 0 & 0 & -3 & & -3 \\ \end{bmatrix}$$

$R_3=R_3+R_1$

$$\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 3 & -6 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 3 & 0 & 0 & -3 & & -3 \\ \end{bmatrix}$$

$R_4=R_4-3R_1$

$$\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 3 & -6 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}$$

$R_2=R_2/3$

$$\begin{bmatrix} 1 & -1 & 2 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}$$

$R_1=R_1+R_2$

$$\begin{bmatrix} 1 &0 & 0 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 1 & -2 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}$$

$R_3=R_3-R_2$

$$\begin{bmatrix} 1 &0 & 0 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 0 & 0 & 0 & & 0 \\ 0 & 3 & -6 & 0 & & 0 \\ \end{bmatrix}$$

$R_4=R_4-3R_2$

$$\begin{bmatrix} 1 &0 & 0 & -1 & & -1 \\ 0 & 1 & -2 & 0 & & 0\\ 0 & 0 & 0 & 0 & & 0 \\ 0 & 0 &0 & 0 & & 0 \\ \end{bmatrix}$$

If $w=t, t\in \R, z=s, s\in \R,$ then $x=-1+t, y=2s, z=s, w=t, t,s\in \R.$

The linear system has infinitely many solutions

$$(-1+t, 2s, s, t), \ t,s\in \R$$