Non trivial solutions !!!!!

Algebra Level 3

If a;b;c are distinct real numbers not equal to one . If ax+y+z=0 x+by+z=0 x+y+cz=0 have a non trivial solution then find the value of 1 1 − a \frac{1}{1-a} + 1 1 − b \frac{1}{1-b} + 1 1 − c \frac{1}{1-c}


The answer is 1.

Abdulmuttalib Lokhandwala by abdulmuttalib lokhandwala , 23

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1 solution

James Wilson
Jan 2, 2021

For the system to have non-trivial solutions, we must have a b c − a − b − c + 2 = 0 , abc-a-b-c+2=0, using the determinant. Solving for c c gives c = a + b − 2 a b − 1 . c=\frac{a+b-2}{ab-1}. Substitute this into 1 1 − a + 1 1 − b + 1 1 − c \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} and simplify: 1 1 − a + 1 1 − b + 1 1 − a + b + 2 a b − 1 \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-\frac{a+b+2}{ab-1}} = 1 1 − a + 1 1 − b + a b − 1 ( 1 − a ) ( 1 − b ) =\frac{1}{1-a}+\frac{1}{1-b}+\frac{ab-1}{(1-a)(1-b)} = 1 − b + 1 − a + a b − 1 ( 1 − a ) ( 1 − b ) =\frac{1-b+1-a+ab-1}{(1-a)(1-b)} = ( 1 − a ) ( 1 − b ) ( 1 − a ) ( 1 − b ) . =\frac{(1-a)(1-b)}{(1-a)(1-b)}.

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