Question

If
$A=\left[\begin{array}{cccc}{e}^t& e^{-t}\cos t& e^{-t}\sin t& \\ e^t& -e^{-t}\cos t-e^{-t}\sin t& -e^{-t}\sin t+e^{-t}\cos t& \\ e^t& 2e^{-t}\sin t& -2e^{-t}\cos t& \end{array}\right]$,

then $A$ is :

• A. invertible only if $t=\pi$.
• B. invertible only if $t=\frac{\pi }{2}$.
• C. invertible for all $t\in R$.
• D. not invertible for any $t\in R$.

Answer 1

The correct option is C invertible for all $t\in R$.

$|A|=\left|\begin{array}{cccc}{e}^t& e^{-t}\cos t& e^{-t}\sin t& \\ e^t& -e^{-t}\cos t-e^{-t}\sin t& -e^{-t}\sin t+e^{-t}\cos t& \\ e^t& 2e^{-t}\sin t& -2e^{-t}\cos t& \end{array}\right|$

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

$R_2\to R_2-R_1,R_3\to R_3-R_1$
$|A|=e^{-t}\left|\begin{array}{ccc}1& \cos t& \sin t\\ 0& -2\cos t-\sin t& -2\sin t+\cos t\\ 0& 2\sin t-\cos t& -2\cos t-\sin t\end{array}\right|$

$R_3\to R_3+2R_2$
$|A|=e^{-t}\left|\begin{array}{ccc}1& \cos t& \sin t\\ 0& -2\cos t-\sin t& -2\sin t+\cos t\\ 0& -5\cos t& -5\sin t\end{array}\right|$

Expanding along $C_1$, we get
$|A|=5e^{-t}\ne 0$
So, $A$ is non-singular.
Hence, matrix $A$ is invertible for all $t\in R$.