Question

Let $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos t& \sin t\\ 0& -\sin t& \cos t\end{array}\right)$
Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the roots of $\text{det}\left(A-\lambda I_3\right)=0$, where $I_3$ denotes the identity matrix. If ${\lambda }_1+{\lambda }_2+{\lambda }_3=\sqrt{2}+1$, then the set of possible values of $t,-\pi \le t<\pi$ is

• A. a void set
• B. $\left\{\frac{\pi }{4}\right\}$
• C. $\left\{-\frac{\pi }{4},\frac{\pi }{4}\right\}$
• D. $\left\{-\frac{\pi }{3},\frac{\pi }{3}\right\}$

Answer 1

Answer: (C)

$\left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& \cos t-\lambda & \sin t\\ 0& -\sin t& \cos t-\lambda \end{array}\right|=0$
$\Rightarrow (1-\lambda )(\cos t-\lambda {)}^2+{\sin }^{2}t=0$
$\Rightarrow (1-\lambda )\left({\lambda }^2-2\lambda \cos t+{\cos }^{2}t\right)+{\sin }^{2}t=0$
$\Rightarrow {\lambda }^2-2x\cos t+{\cos }^{2}t-{\lambda }^3+2{\lambda }^2\cos t+\lambda {\cos }^{2}t+{\sin }^{2}t=0$
$\Rightarrow -{\lambda }^3+{\lambda }^2(1+2\cos t)+\lambda \left({\cos }^{2}t-2\cos t\right)+1=0$
${\lambda }_1+{\lambda }_2+{\lambda }_3=1+2\cos t=1+\sqrt{2}$
$\therefore$ cost $=\frac{1}{\sqrt{2}}$
$\Rightarrow t=\frac{\pi }{4},-\frac{\pi }{4}$