Let A = begin{bmatrix} 1 & 0 & 0 0 & cos t & | vedclass

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(C) The characteristic equation is given by $\det(A - \lambda I_{3}) = 0$. Calculating the determinant: $\begin{vmatrix} 1-\lambda & 0 & 0 \ 0 & \cos

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$\left\{-\frac{\pi}{4}, \frac{\pi}{4}\right\}$

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(C) The characteristic equation is given by $\det(A - \lambda I_{3}) = 0$. Calculating the determinant: $\begin{vmatrix} 1-\lambda & 0 & 0 \ 0 & \cos t-\lambda & \sin t \ 0 & -\sin t & \cos t-\lambda \end{vmatrix} = 0$. Expanding along the first row: $(1-\lambda) [(\cos t - \lambda)^2 - (-\sin^2 t)] = 0$. $(1-\lambda) [\cos^2 t - 2\lambda \cos t + \lambda^2 + \sin^2 t] = 0$. Since $\cos^2 t + \sin^2 t = 1$,we have: $(1-\lambda) [\lambda^2 - 2\lambda \cos t + 1] = 0$. Expanding this: $-\lambda^3 + \lambda^2(1 + 2\cos t) - \lambda(2\cos t + 1) + 1 = 0$. The sum of the roots $\lambda_{1} + \lambda_{2} + \lambda_{3}$ of a cubic equation $a\lambda^3 + b\lambda^2 + c\lambda + d = 0$ is given by $-b/a$. Here,$\lambda_{1} + \lambda_{2} + \lambda_{3} = -\frac{1 + 2\cos t}{-1} = 1 + 2\cos t$. Given $\lambda_{1} + \lambda_{2} + \lambda_{3} = \sqrt{2} + 1$,we equate: $1 + 2\cos t = 1 + \sqrt{2} \Rightarrow 2\cos t = \sqrt{2} \Rightarrow \cos t = \frac{1}{\sqrt{2}}$. For $-\pi \leq t < \pi$,the values of $t$ satisfying $\cos t = \frac{1}{\sqrt{2}}$ are $t = \frac{\pi}{4}$ and $t = -\frac{\pi}{4}$.

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