Question

If a system of equation – ax + y + z = 0, x –by + z = 0

x + y – cz = 0(a, b,c ¹ –1) has a non-zero solution then

$\frac{1}{1 + a}$ + $\frac{1}{1 + b}$ + $\frac{1}{1 + c}$ =

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

D = $\begin{matrix} –a & 1 & 1 \\ 1 & –b & 1 \\ 1 & 1 & –c \end{matrix}$= 0 for non-zero solution

Ž abc – a – b – c – 2 = 0 Ž abc = a + b + c + 2

Now, $\frac{1}{1 + a}$ + $\frac{1}{1 + b}$ + $\frac{1}{1 + c}$

=$\frac{3 + 2(a + b + c) + (ab + bc + ac)}{1 + (a + b + c) + (ab + bc + ac) + abc}$

= $\frac{3 + 2(a + b + c) + (ab + bc + ac)}{1 + 2(a + b + c) + 2 + ab + bc + ac}$ = 1