$$\begin{align} &-3x_1 & &+x_3 &+x_4 &= 0\\ &x_1 &-x_2 &+2x_3 & &= 0\\ & &x_2 &-5x_3 &+2x_4 &= 0\\ &2x_1 & &+2x_3 &-3x_4 &= 0 \end{align} $$

I know the method of doing this by hand and can get the Reduce Row Echelon Form but even then I just get something like,

$x_1 = \frac{5}{8} x_4$

$x_2 = \frac{19}{8} x_4$

$x_3 = \frac{7}{8} x_4$

How do I solve this though? I know the ratios but I just can't seem to figure out what I'm missing. Thank you in advance for your help.

  • $\begingroup$ Check your calculations – or the linear system you posted: I find the matrix of this system is invertible, so there's only the trivial solution., $\endgroup$
    – Bernard
    Apr 23, 2018 at 22:26
  • $\begingroup$ A reduced row-Echelon form should have 1 in the diagonal cells and 0 in the cells for other cells. Not the ratios you have. The reduced row-echelon form should get you the answer for each of the four variables directly. $\endgroup$
    – Leo
    Apr 23, 2018 at 23:07

1 Answer 1


Your answers appear to be correct. You will not be able to find a specific answer for x4. This is because not all 4 equations are linearly independent and in this case, the 4th equation is achieved by adding together a combination of the first 3 equations. In this case, the 4th equation can be achieved: $$Equation4=-(Equation1)-(Equation2)-(Equation3)$$

Hope this helps.

  • $\begingroup$ Which means that the OP is missing one equation, for it to be 4 equations with 4 variables. $\endgroup$
    – Plexus
    Apr 27, 2018 at 12:55

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