Matrices And Determinants Ques 99
The system of equations $\lambda x+y+z=0,-x+\lambda y+z=0$ and $-x-y+\lambda z=0$ will have a non-zero solution, if real values of $\lambda$ are given by …
(1982, 2M)
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Answer:
Correct Answer: 99.$\lambda=0$
Solution:
Formula:
System of equations with 3 variables:
- Given system $\lambda x+y+z=0,-x+\lambda y+z=0$
and $ -x-y+\lambda z=0 $
will have a non-zero solution, if
$ \left|\begin{array}{ccc} \lambda & 1 & 1 \ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{array}\right|=0 $
$\Rightarrow \lambda\left(\lambda^{2}+1\right)-1(-\lambda+1)+1(1+\lambda)=0$
$\Rightarrow \quad \lambda^{3}+3\lambda=0$
$ \Rightarrow \quad \lambda^{3}+3 \lambda=0 $
$ \Rightarrow \quad \lambda\left(\lambda^{2}+3\right)=0 \Rightarrow \lambda=0 \text{ or } \lambda=\pm i\sqrt{3} $
BETA
