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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Determinants

If the system of equations $x+ay+az=0,bx+y+bz=0$ and $cx+cy+z=0$ where $a,b,c$ are non-zero ,non unity has a non-trivial solution then the value of $\large\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$ is

$\begin{array}{1 1}(A)\;-1&(B)\;0\\(C)\;1&(D)\;\large\frac{abc}{a^2+b^2+c^2}\end{array}$

From the given eqation we get,,
x(1-a)=-a(x+y+z)
=> a/(1-a)=-x/(x+y+z) ----(1)
y(1-b)=-b(x+y+z)
=> b/(1-b)=-y/(x+y+z) ----(2)
z(1-c)=-c(x+y+z)
=> c/(1-c)=-z/(x+y+z) -----(3)
adding we get,,
a/(1-a)+b/(1-b)+c/(1-c)=-(x+y+z)/(x+y+z)=-1.

1 Answer

For a non-trivial we must have
$\begin{vmatrix} 1& a& a\\b&1& b\\c& c& 1\end{vmatrix}=0$
Apply $C_1-C_2,C_2-C_3$ we get
$\begin{vmatrix} 1-a& 0& a\\b-1&1-b& b\\0& c-1& 1\end{vmatrix}=0$
$(1-a)[1-b-c(c-1)]+a(b-1)(c-1)=0$
$\large\frac{1}{c-1}+\frac{b}{b-1}+\frac{a}{a-1}=$$0$
$(\large\frac{1}{c-1}$$+1)+\frac{b}{b-1}+\frac{a}{a-1}$$=1$
$\large\frac{c}{c-1}+\frac{b}{b-1}+\frac{a}{a-1}$$=1$
$\large\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}$$=-1$
Hence (A) is the correct answer.
answered Apr 22, 2014 by sreemathi.v
 

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