Matrices And Determinants Ques 3
- If $A=\left[\begin{array}{cc}e^t & e^{-t} \cos t \\ e^t & -e^{-t} \cos t-e^{-t} \sin t \\ e^t & 2 e^{-t} \sin t\end{array}\right.$ $\left.\begin{array}{c}e^{-t} \sin t \\ -e^{-t} \sin t+e^{-t} \cos t \\ -2 e^{-t} \cos t\end{array}\right]$ then $A$ is
(2019 Main, 9 Jan II)
(a) invertible only when $t=\pi$
(b) invertible for every $t \in R$
(c) not invertible for any $t \in R$
(d) invertible only when $t=\frac{\pi}{2}$
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Answer:
Correct Answer: 3.(b)
Solution: (b)
$ \begin{aligned} &|A|=\left|\begin{array}{ccc} 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 & 2 e^{-t} \sin t & -2 e^{-t} \cos t \end{array}\right| \\ &=\left(e^t\right)\left(e^{-t}\right)\left(e^{-t}\right)\left|\begin{array}{ccc} 1 & \cos t & \sin t \\ 1 & -\cos t-\sin t & -\sin t+\cos t \\ 1 & 2 \sin t & -2 \cos t \end{array}\right| \end{aligned} $
(taking common from each column)
Aplying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get
$ \left[\because e^{t-t}=e^0=1\right] $
$=e^{-t}\left|\begin{array}{ccc}1 & \cos t & \sin t \\ 0 & -2 \cos t-\sin t & -2 \sin t+\cos t \\ 0 & 2 \sin t-\cos t & -2 \cos t-\sin t\end{array}\right|$
$=e^{-t}\left((2 \cos t+\sin t)^2+(2 \sin t-\cos t)^2\right)$
(expanding along column 1 )
$=e^{-t}\left(5 \cos ^2 t+5 \sin ^2 t\right)$
$=5 e^{-t}$
$\left(\because \cos ^2 t+\sin ^2 t=1\right)$
$\Rightarrow|A|=5 e^{-t} \neq 0$
for all $t \in R$
$\therefore A$ is invertible for all $t \in R$
$[\because$ If $|A| \neq 0$, then $A$ is invertible $]$
BETA
