Matrices Question 8
Question 8 - 29 January - Shift 2
The set of all values of $t \in \mathbb{R}$, 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} $ invertible, is
(1) ${(2 k+1) \frac{\pi}{2}, k \in \mathbb{Z}}$
(2) ${k \pi+\frac{\pi}{4}, k \in \mathbb{Z}}$
(3) ${k \pi, k \in \mathbb{Z}}$
(4) $\mathbb{R}$
Show Answer
Answer: (4)
Solution:
Formula: Properties of Inverse of a matrix, Determinant properties
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, $R_1 \to R_1-R_2$ then $R_2 \to R_2-R_3$
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,
$e^{-t} \times 1(2 \sin t \cos t+6 \cos ^{2} t+6 \sin ^{2} t-2 \sin t \cos t) \neq 0$
$\Rightarrow e^{-t} \times 6 \neq 0$ $\forall t \in \mathbb{R}$
