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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 \lambda = 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)

best_answer

As we have learned

Homogeneous system of linear equation -

b=0

- wherein

 

 

  1. The given system has a non-trivial solution if \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}\begin{vmatrix} \lambda & \sin \alpha &\cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ -1& \sin \alpha &-\cos \alpha \end{vmatrix}

            On opening the determinant, we get l = sin 2 a + cos 2 a

            For l = 1,          sin 2a + cos 2a = 1       Þ\frac{1}{\sqrt 2 }\sin 2 \alpha + \frac{1}{\sqrt 2} \cos \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 \pi/2 satisfies.

 

Posted by

sudhir.kumar

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