Question

The set of all values of \(t∈R\), for which the matrix \(\begin{bmatrix} e^t & e^{−t}(sin⁡\ t−2cos\ ⁡t) & e^{−t}(−2sin⁡\ t−cos\ ⁡t) \\[0.3em] e^t & e^{−t}(2sin\ ⁡t+cos\ ⁡t) & e^{−t}(sin\ ⁡t−2cos\ ⁡t) \\[0.3em] e^t & e^{−tcos\ ⁡t}& e^{−tsin\ ⁡t} \end{bmatrix}\) is invertible, is

• A. \( \mathbb{R} \)
• B. \( \{(2k+1)\frac \pi2 ,k∈Z\}\)
• C. \(\{k\pi+\frac \pi4 ,k∈Z\}\)
• D. \(\{k\pi,k∈Z\}\)

Answer 1

The Correct Option is A

Step 1: The matrix is:

\( \begin{pmatrix} e^t & e^{-t}( \sin t - 2 \cos t ) & e^{-t}( -2 \sin t + \cos t ) \\ e^t & e^{-t}( 2 \sin t + \cos t ) & e^{-t}( \sin t - 2 \cos t ) \\ e^t & e^{-t} \cos t & e^{-t} \sin t \\ \end{pmatrix} \)

For invertibility, the determinant should be non-zero:

\( \det\left( e^t \cdot e^{-t} \cdot e^{-t} \right) = \det \left( \begin{matrix} \sin t - 2 \cos t & -2 \sin t + \cos t \\ 2 \sin t + \cos t & \sin t - 2 \cos t \\ \cos t & \sin t \\ \end{matrix} \right) \)

This simplifies to:

\( e^t \cdot e^{-t} \cdot e^{-t} \neq 0 \)

Step 2: Applying the row operations: \( R_1 \rightarrow R_1 - R_2 \), \( R_2 \rightarrow R_2 - R_3 \)

We get:

\( e^{-t} \cdot \begin{pmatrix} 0 & -\sin t - \cos t & -3 \sin t + \cos t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \\ \end{pmatrix} \neq 0 \)

Step 3: Now, expand the determinant:

\( e^{-t} \times 1 \left( 2 \sin t \cos t + 6 \cos^2 t + 6 \sin^2 t - 2 \sin t \cos t \right) \neq 0 \)

Conclusion: Therefore, we have:

\( e^{-t} \times 6 \neq 0 \)

Thus, the matrix is invertible for all values of \( t \in \mathbb{R} \).