If A = begin{bmatrix} e^t & e^{-t} cos t & e^ | vedclass

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(C) To check if $A$ is invertible,we calculate its determinant $|A|$.$|A| = e^t \cdot e^{-t} \cdot e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \ 1 & -

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Invertible for all $t \in \mathbb{R}$

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(C) To check if $A$ is invertible,we calculate its determinant $|A|$.$|A| = e^t \cdot e^{-t} \cdot e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \ 1 & -(\cos t + \sin t) & \cos t - \sin t \ 1 & 2 \sin t & -2 \cos t \end{vmatrix} = e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \ 1 & -\cos t - \sin t & \cos t - \sin t \ 1 & 2 \sin t & -2 \cos t \end{vmatrix}$.Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$|A| = e^{-t} \begin{vmatrix} 1 & \cos t & \sin t \ 0 & -2 \cos t - \sin t & \cos t - 2 \sin t \ 0 & 2 \sin t - \cos t & -2 \cos t - \sin t \end{vmatrix}$.Expanding along the first column:$|A| = e^{-t} [(-2 \cos t - \sin t)^2 - (\cos t - 2 \sin t)(2 \sin t - \cos t)]$.$|A| = e^{-t} [4 \cos^2 t + \sin^2 t + 4 \sin t \cos t - (2 \sin t \cos t - \cos^2 t - 4 \sin^2 t + 2 \sin t \cos t)]$.$|A| = e^{-t} [4 \cos^2 t + \sin^2 t + 4 \sin t \cos t + \cos^2 t + 4 \sin^2 t - 4 \sin t \cos t] = e^{-t} [5 \cos^2 t + 5 \sin^2 t] = 5e^{-t}$.Since $5e^{-t} 
eq 0$ for all $t \in \mathbb{R}$,the matrix $A$ is invertible for all $t \in \mathbb{R}$.

If $A = \begin{bmatrix} e^t & e^{-t} \cos t & e^{-t} \sin t \\ e^t & -e^{-t} \cos t - e^{-t} \sin t & -e^{-t} \sin t + e^{-t} \cos t \\ e^t & 2e^{-t} \sin t & -2e^{-t} \cos t \end{bmatrix}$,then $A$ is:

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