Let the homogeneous system of linear equations $$px+y+z=0$$, $$x+qy+z=0$$,$$x+y+rz=0wherep$$,q,r≠$$1have$$ a $$non-trivial$$ solution then $$11$$−$$p+11$$−$$q+11$$−$$r=$$

Question

Let the homogeneous system of linear equations $$px+y+z=0$$, $$x+qy+z=0$$,$$x+y+rz=0wherep$$,q,r≠$$1have$$ a $$non-trivial$$ solution then $$11$$−$$p+11$$−$$q+11$$−$$r=$$

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The system has a $$non-trivial$$ solution⇒$$p111q111r=0apply$$ $$R1$$→$$R1-R3$$, $$R2$$→$$R2-R3$$ ⇒$$p-101-r0q-11-r11r=0$$ ⇒$$p-1q-1r-11-r-0+1-r0-1q-1=0$$ ⇒$$1-p1-qr+1-p1-r+1-q1-r=0$$ Dividing throughout by $$1-p1-q1-r$$, we get          $$r1-r+11-q+11-p=0$$     ⇒ $$r-1+11-r+11-q+11-p=0$$     ⇒$$11-p+11-q+11-r=1$$

Let the homogeneous system of linear equations $p x + y + z = 0, x + q y + z = 0$ , $x + y + r z = 0$ where $p, q, r \neq 1$ have a non-trivial solution then $\frac{1}{1 - p} + \frac{1}{1 - q} + \frac{1}{1 - r} =$

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Let the homogeneous system of linear equations px+y+z=0, x+qy+z=0,x+y+rz=0wherep,q,r≠1have a non-trivial solution then 11−p+11−q+11−r=

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Let the homogeneous system of linear equations $p x + y + z = 0, x + q y + z = 0$ , $x + y + r z = 0$ where $p, q, r \neq 1$ have a non-trivial solution then $\frac{1}{1 - p} + \frac{1}{1 - q} + \frac{1}{1 - r} =$