Help in doing algebraic manipulations. 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.
 A: Notice that (for $x,y,z,x+y+z\ne 0$)
$$(1-p)x=x+y+z\tag{1}$$
$$(1-q)y=x+y+z\tag{2}$$
$$(1-r)z=x+y+z\tag{3}$$
Therefore,

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

Note that the zero determinant condition is built in to this:
$\frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=1\Leftrightarrow3-2(p+q+r)+pr+qr+pq=(1-p)(1-q)(1-r)$
$\Leftrightarrow p+q+r=2+pqr$,
Which is exactly what we get from $det A=0$ for the coefficient matrix.
A: 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.
Suppose
$\frac{1}{1-p}+\frac{1}{1-q}+\frac{1}{1-r}=k;$
then, by multiplying out and distributing, etc, we'll get that:
$3-2(p+q+r)+pq+pr+qr=k-k(p+q+r)+k(pq+pr+qr)-kpqr.$
Substituting that $pqr=p+q+r-2$, we get:
$3-2(p+q+r)+pq+pr+qr=3k-2k(p+q+r)+k(pq+pr+qr).$
This equation can then be factored:
$(1-k)(3-2(p+q+r)+pq+pr+qr)=0.$
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.
