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The Eigen values of matrix are:

a) ± cos∝

b) ± sin ∝

c) tan ∝  & cot ∝

d) cos ∝  ±  sin ∝

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option D)

The equation is $(\cos \alpha -\lambda )^{2}-\sin ^{2}\alpha =0$

which is a quadratic equation in $\lambda$.

Solve using sridhara acharya's formula. You get option D
by Active (1.5k points)
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Can u comment the formula here..

Wil be useful for all!!
This is the sridhara acharya's formula which all of us had studied in our 10th standard :p

$x=(-b\pm \sqrt{b^{2}-4ac})/(2*a)$
where a,b,c are the constants of the general form of a quadratic equation ax^2+bx+c=0
by Active (1.5k points)

sum of diagnals element = sum of eigen values

by Boss (36.9k points)

Let A be the matrix given in the question.

Now,Trace(A)=sum of eigen values of A

or,Trace(A)=cos α + cos α=2*cos α

Now,by adding and subtracting sin α we get:

or,2cos α=(cos α + sin α) + (cos α - sin α)= λ1 + λ 2

Therefore, 2 eigen values can be denoted by (cos ∝  ±  sin ∝ (d))

So, Option (d) is the ans.

by Junior (909 points)