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Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \text { cost } & \text { sint } \\ 0 & -\sin t & \cos t\end{array}\right)$
Let $\lambda_{1}, \lambda_{2}, \lambda_{3}$ be the roots of $\operatorname{det}\left(A-\lambda I_{3}\right)=0$, where $I_{3}$ denotes the identity matrix. If $\lambda_{1}+\lambda_{2}+\lambda_{3}=\sqrt{2}+1$, then the set of possible values of $t,-\pi \leq t < \pi$ is
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Let $\lambda_{1}, \lambda_{2}, \lambda_{3}$ be the roots of $\operatorname{det}\left(A-\lambda I_{3}\right)=0$, where $I_{3}$ denotes the identity matrix. If $\lambda_{1}+\lambda_{2}+\lambda_{3}=\sqrt{2}+1$, then the set of possible values of $t,-\pi \leq t < \pi$ is
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The correct answer is:
$\left\{-\frac{\pi}{4}, \frac{\pi}{4}\right\}$
$\left|\begin{array}{ccc}1-\lambda & 0 & 0 \\ 0 & \cos t-\lambda & \sin t \\ 0 & -\sin t & \cos t-\lambda\end{array}\right|=0$
$\Rightarrow(1-\lambda)(\cos t-\lambda)^{2}+\sin ^{2} t=0$
$\Rightarrow(1-\lambda)\left(\lambda^{2}-2 \lambda \cos t+\cos ^{2} t\right)+\sin ^{2} t=0$
$\Rightarrow \lambda^{2}-2 x \cos t+\cos ^{2} t-\lambda^{3}+2 \lambda^{2} \cos t+\lambda \cos ^{2} t+\sin ^{2} t=0$
$\Rightarrow-\lambda^{3}+\lambda^{2}(1+2 \cos t)+\lambda\left(\cos ^{2} t-2 \cos t\right)+1=0$
$\lambda_{1}+\lambda_{2}+\lambda_{3}=1+2 \operatorname{cost}=1+\sqrt{2}$
$\therefore$ cost $=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{t}=\frac{\pi}{4},-\frac{\pi}{4}$
$\Rightarrow(1-\lambda)(\cos t-\lambda)^{2}+\sin ^{2} t=0$
$\Rightarrow(1-\lambda)\left(\lambda^{2}-2 \lambda \cos t+\cos ^{2} t\right)+\sin ^{2} t=0$
$\Rightarrow \lambda^{2}-2 x \cos t+\cos ^{2} t-\lambda^{3}+2 \lambda^{2} \cos t+\lambda \cos ^{2} t+\sin ^{2} t=0$
$\Rightarrow-\lambda^{3}+\lambda^{2}(1+2 \cos t)+\lambda\left(\cos ^{2} t-2 \cos t\right)+1=0$
$\lambda_{1}+\lambda_{2}+\lambda_{3}=1+2 \operatorname{cost}=1+\sqrt{2}$
$\therefore$ cost $=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{t}=\frac{\pi}{4},-\frac{\pi}{4}$
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