Question

If the system of equations

x + ay + az = 0

bx + y + bz = 0

cx + cy + z = 0

where a, b and c are non-zero and non-unity, has a non-trivial solution, then the value of $\frac{a}{1–a}$+ $\frac{b}{1–b}$ + $\frac{c}{1–c}$ is –

• A. Zero
• B. 1
• C. – 1
• D. $\frac{abc}{a^{2} + b^{2} + c^{2}}$

Answer 1

Answer: (C)

– 1

Answer 2

For non trivial solution ⇒ ∆ = 0

∆=$\left| \begin{matrix} 1 & a & a \\ b & 1 & b \\ c & c & 1 \end{matrix} \right|$=∆=$\left| \begin{matrix} 1–a & 0 & a \\ b–1 & 1–b & b \\ c & c–1 & 1 \end{matrix} \right|$=0⇒n(1–a)(1–b)(1–c)× $\left| \begin{matrix} 1 & 0 & \frac{a}{1–a} \\ –1 & 1 & \frac{b}{1–b} \\ 0 & –1 & \frac{1}{1–c} \end{matrix} \right|$ = 0⇒ 1/1–a + b/1–b + c/1–c = – 1