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The system of equations $\lambda x+y+z=0$, $-x+\lambda y+z=0$, $-x-y+\lambda z=0$ will have a non-zero solutions if real values of $\lambda$ are given by

(A) $\lambda=1$ (B) $\lambda=2$ (C) $\lambda=0$ (D) $\lambda=4$

1 Answer

The given homogeneous system of equations will have non zero solution if $D=0$
$\Rightarrow \begin{vmatrix}\lambda &1 &1\\-1 &\lambda&1\\-1&-1&\lambda\end{vmatrix}=0$
$\Rightarrow \lambda(\lambda^2+1)-1(-\lambda+1)+1(1+\lambda)=0$
$\Rightarrow \lambda^3+3\lambda=0$
$\Rightarrow \lambda(\lambda^2+3)=0$
But $\lambda^2+3\neq 0$ for real $\lambda$
$\Rightarrow \lambda=0$
Hence (c) is the correct answer.
answered Nov 19, 2013 by sreemathi.v

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