4
$\begingroup$

$x_1 - x_2 - 2x_3 + x_4 = 0 \\ -3x_1 + 3x_2 + x_3 - x_4 = 0 \\ 2x_1 - 2x_2 + x_3 = 0$

How do I solve this system of equations? I know this is a homogenous system. By applying elementary row operations, I get the following:

$x_1 - x_2 + 1/5x_4 = 0$

$x_3 - \frac{2}{5}x_4 = 0$

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Since the homogenous linear system possesses more unknowns than equations $(n > m)$, then it will have nontrivial solutions.

In this case, you can just assign arbituary values to $x_2, x_4$

$x_1 = a - \frac{1}{5}b$

$x_2 = a$

$x_3 = \frac{2}{5}b$

$x_4 = b$

Which can be made even simpler by replacing $b$ with $5b^{'}$

$x_1 = a - b^{'}$

$x_2 = a$

$x_3 = 2b^{'}$

$x_4 = 5b^{'}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .