What is the fundamental solution of this homogeneous system of linear equations? And what are the steps to solving one? $$\left\{ \begin{array}{ll} x_1+3x_2-5x_3+9x_4-x_5=0 \\ 2x_1-2x_2-3x_3-7x_4+2x_5=0 \\ x_1-5x_2+2x_3-16x_4+3x_5=0 \end{array} \right.$$

I had this exact task in a test and couldn't wrap my mind around it. I'd appreciate a more or less simple answer as I'm not a native English speaker. Thanks in advance.

  • $\begingroup$ Would the answer to ''finding a fundamental solution'' be - there is none, because there is an infinite amount of solutions? Or should I have x3=c3; x4=c4; x5=c5 and then make a solution out of that? Meaning x2=7/8*c3 - 25/8*c4+1/2*c5 etc.? I'm wondering if that counts as a fundamental solution $\endgroup$ Commented Jan 26, 2017 at 21:21
  • $\begingroup$ Thanks a lot for your help :) $\endgroup$ Commented Jan 26, 2017 at 21:30

1 Answer 1


Since the right side of each equation is 0, it is trivial that $x_1= x_2= x_3= x_4= x_5= 0$ is a solution. But since there are three equations in five unknowns, there exist other, non trivial solutions.

One way to find all possible solutions is to start trying to solve the equations. For example, if we multiply the first equation by 2 and subtract from the second equation we eliminate both $x_1$ and $x_5$: $-8x_2+ 7x_3- 25x_4= 0$. If we multiply the third equation by 2 and subtract from the second equation, we eliminate $x_1$ again: $14x_2- 7x_3+ 25x_4- 4x_5= 0$. And if we add those two equations, we eliminate both $x_3$ and $x_4$: 6x_2- 4x_5= 0. From that $x_5= \frac{3}{2}x_2$. Putting that into $14x_2- 7x_3+ 25x_4- 4x_5= 0$ we get $8x_2- 7x_3+ 25x_1= 0$. Solving that for $x_1$, $x_1= \frac{7}{25}x_3- \frac{8}{25}x_2$. We are through with the equations so this is as far as we can go. writing "A" for $x_3$ and "B" for $x_2$, $x_1= \frac{7}{25}A- \frac{8}{25}B$, $x_2= A$, $x_4= B$, and you can solve for $x_3$ and $x_5$ in terms of A and B.


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