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Let \lambda and \alpha be real. Find the set of all values of \lambda for which the system of linear equations

            \lambdax + sin\alpha .y + cos\alpha . z = 0

            x + cos\alpha . y  + sin\alpha . z = 0

            -x+ sin\alpha . y-cos\alpha . z = 0 has a non-trivial solution. For l = 1, The possible values of \alpha are

Option: 1

\pi /8


Option: 2

\pi /6


Option: 3

\pi /2


Option: 4

\pi /3


Answers (1)

As we have learned

Homogeneous system of linear equation -

b=0

- wherein

The given system has a non-trivial solution if

\begin{vmatrix} \lambda &\sin \alpha & \cos \alpha \\ 1 & \cos \alpha &\sin \alpha \\ -1&\sin \alpha & -\cos \alpha \end{vmatrix}=0

            On opening the determinant, we get \lambda = sin 2 \alpha + cos 2 \alpha

            For \lambda = 1,          sin 2\alpha + cos 2\alpha = 1     \Rightarrow \frac{1}{\sqrt{2}}\sin 2\alpha +\frac{1}{\sqrt{2}}\cos 2\alpha =\frac{1}{\sqrt{2}}

           

            \cos \left ( 2\alpha -\frac{\pi }{4} \right )=\frac{1}{\sqrt{2}}=\cos \left ( 2n\pi \pm \frac{\pi }{4} \right )\Rightarrow 2\alpha =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{4}, Hence satisfies.\pi /2

 

 

Posted by

Kshitij

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