Let the homogeneous system of linear equations $px+y+z=0$, $x+qy+z=0$, $x+y+rz=0$ where $p,q,r$ are not equal to $1$, have a non zero solution, then the value of $\frac{1}{1-p} + \frac{1}{1-q} + \frac{1}{1-r}$ is?

I know that because this a homogeneous system the condition implies that the determinant of the coefficient matrix vanishes. But I am having trouble calculating the given value. Any help would be appreciated.


Notice that (for $x,y,z,x+y+z\ne 0$)

$$(1-p)x=x+y+z\tag{1}$$ $$(1-q)y=x+y+z\tag{2}$$ $$(1-r)z=x+y+z\tag{3}$$


$$ \frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=1. $$

Note that the zero determinant condition is built in to this:


$\Leftrightarrow p+q+r=2+pqr$,

Which is exactly what we get from $det A=0$ for the coefficient matrix.

  • $\begingroup$ smooth. You didn't use the vansihing determinant condition. $\endgroup$ – user34304 Apr 29 '15 at 15:38
  • $\begingroup$ @user34304, Glad you like it. ;) $\endgroup$ – ki3i Apr 29 '15 at 15:49

As elegant as @ki3i's solution is, what if we weren't lucky enough to see the easy way out?

We'd have to start from the constraint given by the determinant that $p+q+r=2+pqr$ and make it look like the desired expression.



then, by multiplying out and distributing, etc, we'll get that:


Substituting that $pqr=p+q+r-2$, we get:


This equation can then be factored:


Here, we can clearly see that if $k=1$ we're done. So it's clear that the expression can evaluate to one.

To finish things off and show that it must evaluate to 1, we should show that the second factor is 0 and the determinant constraint holds $\Leftrightarrow$ (at least) one of $p,q,r$ is 1.

(completed by @ki3i) Since $$ \begin{align} 0&=&3-2(p+q+r)+pq+pr+qr \\&=&3-(p+q+r)-(p+q+r)+pq+pr+qr\\ &\stackrel{det=0}{=}&3-(2+pqr)-(p+q+r)+pq+pr+qr\\ &=&1-p-q-r+pq+pr+qr-pqr\\ &=&(1-p)(1-q)(1-r)\, \end{align} $$ iff $p=1$ or $q=1$ or $r=1$, this completes the proof.

  • 1
    $\begingroup$ thanks for the edit,...thought I should return the favour. ;) $\endgroup$ – ki3i Apr 30 '15 at 17:06
  • $\begingroup$ thanks! I thought it should factor to that but the algebra was being unfriendly $\endgroup$ – MichaelChirico Apr 30 '15 at 17:11

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