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

If the system of equations $ax+y+z=0$, $x+by+z=0$, $x+y+cz=0$, where $(a,b,c\neq 1)$ has a non-trivial solutions, then the value of $\large\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is

$(a)\;-1\qquad(b)\;0\qquad(c)\;1\qquad(d)\;None\;of\;these$

1 Answer

For given homogeneous equations to have non-zero solutions,
$\begin{vmatrix}a&1&1\\1&b&1\\1&1&c\end{vmatrix}=0$
$\begin{vmatrix}a&1&1\\1-a&b-1&0\\1-a&0&c-1\end{vmatrix}=0$
$\Rightarrow a(b-1)(c-1)-(1-a)(c-1)-(1-a)(b-1)=0$
Dividing by $(1-a)(1-b)(1-c)$ we get,
$\large\frac{a}{1-a}+\frac{1}{1-c}+\frac{1}{1-b}$$=0$
$\large\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$$=1$
Hence (c) is the correct answer.
answered Nov 26, 2013 by sreemathi.v
 

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