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Let $\lambda_1, \lambda_2, \lambda_3$ denote the eigenvalues of the matrix

$$A \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos t & \sin t \\ 0 & – \sin t & \cos t \end{pmatrix}.$$ If $\lambda_1+\lambda_2+\lambda_3 = \sqrt{2}+1$, then the set of possible values of $t, \: – \pi \leq t < \pi$, is

  1. Empty set
  2. $\{ \frac{\pi}{4} \}$
  3. $\{ – \frac{\pi}{4}, \frac{\pi}{4} \}$
  4. $\{ – \frac{\pi}{3}, \frac{\pi}{3} \}$
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2 Answers

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1 votes
We know that   $T_{r}\left ( A \right )$  $=$  $Trace\left ( A \right )$  $=$  sum of principle diagonal elements  $=$ sum of eigen values of the matrix.

$\therefore$  $2$$cost$  $+$  $1$  $=$  $\sqrt{2}$  $+$ $1$

$\Rightarrow$   $cost$  $=$  $\frac{1}{\sqrt{2}}$  

$\Rightarrow$    $t$ $=$ $-$ $\frac{\pi }{4}$  or  $\frac{\pi }{4}$  (for the interval  $\left [ -\pi ,\pi \right ]$ ,  $cosx$ is positive between $\left [ -\frac{\pi }{2},\frac{\pi}{2} \right ]$).

$\therefore$  Option C is the correct answer.
1 votes
1 votes
Ans is C.

The sum of eigen values of any matrix = sum of the main diagonal elements of matrix.

So, $2\cos t + 1=\sqrt{2}+1$

so t=-π/4 or π/4

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