Question

The set of all values of $t \in R$, for which the matrix
$ \begin{bmatrix} 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{bmatrix} $ is invertible, is

• A. $\left\{(2 k +1) \frac{\pi}{2}, k \in Z \right\}$
• 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 $\neq 0$
So,
$\begin{vmatrix}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t\end{vmatrix} \neq 0$
$\Rightarrow e ^{ t } \cdot e ^{- t } \cdot e ^{- t }\begin{vmatrix}1 & \sin t -2 \cos t & -2 \sin t -\cos t \\ 1 & 2 \sin t +\cos t & \sin t -2 \cos t \\ 1 & \cos t & \sin t\end{vmatrix} \neq 0$
Applying, $R_1 \rightarrow R_1-R_2$ then $R_2 \rightarrow R_2-R_3$
We get
$e ^{- t }\begin{vmatrix}0 & -\sin t-\cos t & -3 \sin t+\cos t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t\end{vmatrix} \neq 0$
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) \neq 0$
$\Rightarrow e ^{-t} \times 6 \neq 0$
$ \text { for } \forall t \in R$