Answer:
The system to solve is
[tex]x_1-2x_2+x_3+4x_4-2x_5=0\\-2x1+4x_2+x_3-2x_4-4x_5=0\\3x_1-6x_2+8x_3+4x_4-13x_5=0\\8x_1-16x_2+7x_3+12x_4-6x_5=0[/tex]
and the augmented matrix of the system is [tex]\left[\begin{array}{cccccc}1&-2&1&4&-2&0\\-2&4&1&-2&-4&0\\3&-6&8&4&-13&0\\8&-16&7&12&-6&0\end{array}\right][/tex]
Using row operations we obtain the echelon form of the matrix, that is,
[tex]\left[\begin{array}{ccccccc}1&-2&1&4&-2&0\\0&0&-1&-20&10&0\\0&0&0&-54&22&0\\0&0&0&0&13&0\end{array}\right][/tex]
Now, we use backward substitution:
1. [tex]13x_5=0,\; x_5=0[/tex]
2.
[tex]54x_4+22x_5=0\\54x_4+22*0=0\\x_4=0[/tex]
3.
[tex]-x_3-20x_4+10x_5=0\\-x_3-20*0+10*0=\\x_3=0[/tex]
4.
[tex]x_1-2x_2+x_3+4x_4-2x_5=0\\x_1-2x_2+0+4*0-2*0=0\\x_1=2x_2[/tex]
The system has infinite solutions and the set of solutions is:
[tex]\{(x_1,x_2,x_3,x_4,x_5)=(2t,t,0,0,0): t\in\mathbb{R}\}[/tex]
