Mathematics - Matrices Question with Solution | TestHub

Let

$A=\left[\begin{array}{ccc}{\mathrm{e}}^t& {\mathrm{e}}^{-t}\left(\sin t-2\cos t\right)& {\mathrm{e}}^{-t}\left(-2\sin t-\cos t\right)\\ {\mathrm{e}}^t& {\mathrm{e}}^{-t}\left(2\sin t+\cos t\right)& {\mathrm{e}}^{-t}\left(\sin t-2\cos t\right)\\ {\mathrm{e}}^t& {\mathrm{e}}^{-t}\cos t& {\mathrm{e}}^{-t}\sin t\end{array}\right]$

Given matrix $A$ is invertible if

$\left|A\right|\ne 0$

$\Rightarrow \left|\begin{array}{c}\begin{array}{ccc}{\mathrm{e}}^t& {\mathrm{e}}^{-t}\left(\sin t-2\cos t\right)& {\mathrm{e}}^{-t}\left(-2\sin t-\cos t\right)\\ {\mathrm{e}}^t& {\mathrm{e}}^{-t}\left(2\sin t+\cos t\right)& {\mathrm{e}}^{-t}\left(\sin t-2\cos t\right)\\ {\mathrm{e}}^t& {\mathrm{e}}^{-t}\cos t& {\mathrm{e}}^{-t}\sin t\end{array}\end{array}\right|\ne 0$

$\Rightarrow e^t·e^{-t}·e^{-t}\left|\begin{array}{ccc}1& \sin t-2\cos t& -2\sin t-\cos t\\ 1& 2\sin t+\cos t& \sin t-2\cos t\\ 1& \cos t& \sin t\end{array}\right|\ne 0$

Applying ${\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_2$ then ${\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3$, ee get

$e^{-t}\left|\begin{array}{ccc}0& -\sin t-3\cos t& -3\sin t+\cos t\\ 0& 2\sin t& -2\cos t\\ 1& \cos t& \sin t\end{array}\right|\ne 0$

By expanding we have,

$e^{-t}\times \left(2\sin t\mathrm{co}st+6{\cos }^{2}t+6{\sin }^{2}t-2\sin t\cos t\right)\ne 0$

$\Rightarrow e^{-t}\times 6\ne 0$ for $\forall t\in \mathrm{ℝ}$