If the system of equations $$ax+y+z=0$$, $$x+by+z=0$$, $$x+y+cz=0$$ $$(a$$,b,c≠$$1)has$$ a $$non-zero$$ solution. $$Then11$$−$$a+11$$−$$b+11$$−$$c=$$

Question

If the system of equations  $$ax+y+z=0$$,  $$x+by+z=0$$,  $$x+y+cz=0$$  $$(a$$,b,c≠$$1)has$$ a $$non-zero$$ solution. $$Then11$$−$$a+11$$−$$b+11$$−$$c=$$

Infinity Learn · Accepted Answer

The condition for a homogeneous system of equations to have a $$non-zero$$ solution is Δ$$=0$$⇒$$a111b111c=0$$      $$R3$$→$$R3-R2$$, $$R2$$→$$R2-R1$$ ⇒$$a111-ab-1001-bc-1=0$$ ⇒$$ab-1c-1-11-ac-1+11-a1-b=0$$ ⇒$$a1-b1-c+11-a1-c+11-a1-b=0$$ ⇒$$a1-a+11-b+11-c=0$$   dividing through out by $$1-a1-b1-c$$ ⇒$$1+a-11-a+11-b+11-c=0$$⇒$$11$$−$$a+11$$−$$b+11$$−$$c=1$$

If the system of equations $a x + y + z = 0, x + b y + z = 0, x + y + c z = 0 (a, b, c \neq 1)$ has a non-zero solution. Then $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} =$

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If the system of equations ax+y+z=0, x+by+z=0, x+y+cz=0 (a,b,c≠1)has a non-zero solution. Then11−a+11−b+11−c=

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If the system of equations $a x + y + z = 0, x + b y + z = 0, x + y + c z = 0 (a, b, c \neq 1)$ has a non-zero solution. Then $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} =$