Question

If $A = \begin{bmatrix}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{bmatrix} $ Then $A$ is -

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

Answer 1

The Correct Option is C

To determine whether the matrix \(A\) is invertible for given values of \(t\), we need to check whether the determinant of the matrix is non-zero.

The given matrix \(A\) is:

\(e^{t}\)	\(e^{t} \cos t\)	\(e^{-t}\sin t\)
\(e^{t}\)	\(-e^{t} \cos t -e^{-t}\sin t\)	\(-e^{-t} \sin t+ e^{-t} \cos t\)
\(e^{t}\)	\(2e^{-t} \sin t\)	\(-2e^{-t} \cos t\)

For a matrix to be invertible, its determinant must be non-zero. We will calculate the determinant of the given 3x3 matrix by expanding along the first row:

\(\text{det}(A) = e^t \left((-e^{t} \cos t - e^{-t}\sin t)(-2e^{-t} \cos t) - (2e^{-t} \sin t)(-e^{-t} \sin t+ e^{-t} \cos t)\right)\)

On simplifying the determinant expression:

1. Calculate the expansion components:

• \((-e^{t} \cos t - e^{-t}\sin t)(-2e^{-t} \cos t) = 2 \cos^2 t + 2 \sin t \cos t\)
• \((-2e^{-t} \sin t)e^{-t}(\cos t - \sin t) = -2 \sin t (\cos t - \sin t) = -2 \sin t \cos t + 2\sin^2 t\)

2. Add these components to find the determinant:

\(\text{det}(A) = e^t(2 \cos^2 t + 2 \sin t \cos t + 2 \sin t \cos t - 2 \sin^2 t)\)

\(\text{det}(A) = e^t(2 (\cos^2 t - \sin^2 t))\)

\(\text{det}(A) = e^t(2 \cos(2t))\)

The determinant is \(e^t(2 \cos(2t))\), which is non-zero for all real \(t\) except when \(\cos(2t) = 0\). However, because \(e^t\) is never zero, \(\text{det}(A)\) is non-zero for all \(t \in \mathbb{R}\).

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

The correct answer is: invertible for all \(t \epsilon R\).