College

Solve the following system of equations by performing Gaussian Elimination on the corresponding augmented matrix:

\[
\begin{align*}
x_1 - 2x_2 + x_3 + 4x_4 - 25 &= 0 \\
-2x_1 + 4x_2 + x_3 - 224 - 45 &= 0 \\
3x_1 - 6x_2 + 8x_3 + 4x_4 - 13x_5 &= 0 \\
8x_1 - 16x_2 + 7x_3 + 12x_4 - 6x_5 &= 0
\end{align*}
\]

Answer :

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]