Question

Assertion :If the system of equations $x+ay+az=0,bx+y+bz=0$ & $cx+cy+z=0$ where a, b, c are non zero non unity, has a non trivial solution, then value of $\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$ is 1 Reason: For non trivial solution determinant of coefficient matrix is,zero,

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A),
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A)is true but (R} is false,
• D. (A)is false but (R} is true.

Answer 1

The correct option is D (A)is false but (R} is true.

For non trivial solution we must have $\left|\begin{array}{ccc}1& a& a\\ b& 1& b\\ c& c& 1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1-a& 0& a\\ b-1& 1-b& b\\ 0& c-1& 1\end{array}\right|=0$
applying $C_1\to C_1-C_2,C_2\to C_2-C_3$
$\Rightarrow (1-a)(1-b)-b(c-1)]+a[(b-1)(c-1)]=0$
$\Rightarrow (1-a)(1-b)+b(1-a)(1-c)+a(1-b)(1-c)=0$
$\Rightarrow \frac{1}{1-c}+\frac{b}{1-b}+\frac{a}{1-a}=0$
$\Rightarrow \frac{1}{1-c}-1+\frac{b}{1-b}+\frac{a}{1-a}=-1$
$\Rightarrow \frac{c}{1-c}+\frac{b}{1-b}+\frac{a}{1-a}=-1$
$\therefore$ Assertion (A) is false & Reason (R) is true.
Hence, option D.