If a system of equation - ax + y + z = 0 x - by + z = 0 x + y - cz = 0 (a, b, c ≠ -1) has a non-zero solution then (1/1+a) + (1/1+b) + (1/1+c) =

Question

If a system of equation - ax + y + z = 0 x - by + z = 0 x + y - cz = 0 (a, b, c ≠ -1) has a non-zero solution then (1/1+a) + (1/1+b) + (1/1+c) =  (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Δ = |-a&1&1 1&-b&1 1&1&-c| = 0 for non-zero solution ⇒ abc - a - b - c - 2 = 0 ⇒ abc = a + b + c + 2 Now, (1/1+a) + (1/1+b) + (1/1+c) = (3+2(a+b+c)+(ab+bc+ac)/1+(a+b+c)+(ab+bc+ac)+abc) = (3+2(a+b+c)+(ab+bc+ac)/1+2(a+b+c)+2+ab+bc+ac) = 1

Q. If a system of equation −ax+y+z=0 x−by+z=0 x+y−cz=0(a,b,c=−1) has a non-zero solution then 1+a1​+1+b1​+1+c1​=

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Δ=∣∣​−a11​1−b1​11−c​∣∣​=0 for non-zero solution ⇒abc−a−b−c−2=0 ⇒abc=a+b+c+2 Now, 1+a1​+1+b1​+1+c1​ =1+(a+b+c)+(ab+bc+ac)+abc3+2(a+b+c)+(ab+bc+ac)​ =1+2(a+b+c)+2+ab+bc+ac3+2(a+b+c)+(ab+bc+ac)​=1

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