Question

If the system of equations ax + y + z = 0,

x + by + z = 0 and x + y + cz = 0 (a, b, c ¹ 1) has a non-trivial

solution, then the value of $\frac{1}{1–a}$ + $\frac{1}{1–b}$ + $\frac{1}{1–c}$ is –

• A. –1
• B. 0
• C. 1
• D. None of these

Answer 1

Answer: (C)

1

Answer 2

Non trivial sol D = 0

Ž$\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix} \right|$ = 0

$\begin{matrix} C_{1} \rightarrow C_{1} - C_{2} \\ C_{2} \rightarrow C_{2} - C_{3} \end{matrix} \Rightarrow \left| \begin{matrix} a - 1 & 0 & 1 \\ 1 - b & b - 1 & 1 \\ 0 & 1 - c & c \end{matrix} \right| = 0$ Ž (1 – a) (1 – b)

(1 – c) $\left| \begin{matrix}

• 1 & 0 & \frac{1}{1 - a} \ 1 & - 1 & \frac{1}{1 - b} \ 0 & 1 & \frac{c}{1 - c} \end{matrix} \right| = 0$

Ž (1 – a) (1 – b) (1 – c) $\left\lbrack \frac{1}{1 - a}(1)–\frac{1}{1 - b}( - 1) + \frac{c}{1 - c} \right\rbrack$ = 0

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