Question

The set of all values of $t\in R$, for which the matrix
$\left[\begin{array}{ccc}{e}^t& e^{-t}(\sin t-2\cos t)& e^{-t}(-2\sin t-\cos t)\\ e^t& e^{-t}(2\sin t+\cos t)& e^{-t}(\sin t-2\cos t)\\ e^t& e^{-t}\cos t& e^{-t}\sin t\end{array}\right]$ is invertible, is

• A. $\left\{(2k+1)\frac{\pi }{2},k\in Z\right\}$
• B. $\left\{k\pi +\frac{\pi }{4},k\in Z\right\}$
• C. $\{k\pi ,k\in Z\}$
• D. $R$

Answer 1

Answer: (D)

If its invertible, then determinant value $\ne 0$
So,
$\left|\begin{array}{ccc}{e}^t& e^{-t}(\sin t-2\cos t)& e^{-t}(-2\sin t-\cos t)\\ e^t& e^{-t}(2\sin t+\cos t)& e^{-t}(\sin t-2\cos t)\\ e^t& e^{-t}\cos t& e^{-t}\sin t\end{array}\right|\ne 0$
$\Rightarrow e^t\cdot e^{-t}\cdot e^{-t}\left|\begin{array}{ccc}1& \sin t-2\cos t& -2\sin t-\cos t\\ 1& 2\sin t+\cos t& \sin t-2\cos t\\ 1& \cos t& \sin t\end{array}\right|\ne 0$
Applying, $R_1\to R_1-R_2$ then $R_2\to R_2-R_3$
We get
$e^{-t}\left|\begin{array}{ccc}0& -\sin t-\cos t& -3\sin t+\cos t\\ 0& 2\sin t& -2\cos t\\ 1& \cos t& \sin t\end{array}\right|\ne 0$
By expanding we have,
$e^{-t}\times 1\left(2\sin t\cos t+6{\cos }^{2}t+6{\sin }^{2}t-2\sin t\cos t\right)\ne 0$
$\Rightarrow e^{-t}\times 6\ne 0$
$\text{ for }\forall t\in R$