1. Key Concept: Invertibility of a Matrix

A square matrix $A$ is invertible (or non-singular) if and only if its determinant, denoted as $\det(A)$ or $|A|$, is non-zero. If $\det(A) = 0$, the matrix is singular and not invertible. Our goal is to find the set of all values of $t \in \mathbb{R}$ for which the given matrix has a non-zero determinant.

2. Setting up the Determinant

Let the given matrix be $A$. A = \left[ {\begin{matrix} {{e^t}} & {{e^{ - t}}(\sin t - 2\cos t)} & {{e^{ - t}}( - 2\sin t - \cos t)} \cr {{e^t}} & {{e^{ - t}}(2\sin t + \cos t)} & {{e^{ - t}}(\sin t - 2\cos t)} \cr {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr } } \right]

3. Simplifying the Determinant using Column Operations

To simplify the calculation of the determinant, we can factor out common terms from columns.

• Factor out $e^t$ from the first column ($C_1$).
• Factor out $e^{-t}$ from the second column ($C_2$).
• Factor out $e^{-t}$ from the third column ($C_3$).

The property of determinants states that if a column of a matrix is multiplied by a scalar $k$, the determinant is multiplied by $k$. So, \det(A) = e^t \cdot e^{-t} \cdot e^{-t} \cdot \det \left[ {\begin{matrix} 1 & {(\sin t - 2\cos t)} & {(- 2\sin t - \cos t)} \cr 1 & {(2\sin t + \cos t)} & {(\sin t - 2\cos t)} \cr 1 & {\cos t} & {\sin t} \cr } } \right] This simplifies to: \det(A) = e^{-t} \cdot \det \left[ {\begin{matrix} 1 & {\sin t - 2\cos t} & -2\sin t - \cos t \cr 1 & {2\sin t + \cos t} & \sin t - 2\cos t \cr 1 & {\cos t} & \sin t \cr } } \right] Let's call the $3 \times 3$ matrix inside the determinant $A'$. So, $\det(A) = e^{-t} \det(A')$.

4. Further Simplification using Row Operations

To make the determinant calculation easier, we can create zeros in the first column of $A'$. This is achieved by performing row operations:

• $R_2 \to R_2 - R_1$ (Subtract Row 1 from Row 2)
• $R_3 \to R_3 - R_1$ (Subtract Row 1 from Row 3)

These row operations do not change the value of the determinant. Let's calculate the new elements for the second and third rows:

For $R_2 \to R_2 - R_1$:

• New $A'_{21} = 1 - 1 = 0$
• New $A'_{22} = (2\sin t + \cos t) - (\sin t - 2\cos t) = 2\sin t + \cos t - \sin t + 2\cos t = \sin t + 3\cos t$
• New $A'_{23} = (\sin t - 2\cos t) - (-2\sin t - \cos t) = \sin t - 2\cos t + 2\sin t + \cos t = 3\sin t - \cos t$

For $R_3 \to R_3 - R_1$:

• New $A'_{31} = 1 - 1 = 0$
• New $A'_{32} = \cos t - (\sin t - 2\cos t) = \cos t - \sin t + 2\cos t = 3\cos t - \sin t$
• New $A'_{33} = \sin t - (-2\sin t - \cos t) = \sin t + 2\sin t + \cos t = 3\sin t + \cos t$

Now, the determinant of $A'$ becomes: \det(A') = \det \left[ {\begin{matrix} 1 & {\sin t - 2\cos t} & -2\sin t - \cos t \cr 0 & {\sin t + 3\cos t} & 3\sin t - \cos t \cr 0 & {3\cos t - \sin t} & 3\sin t + \cos t \cr } } \right]

5. Expanding the Determinant

We can now expand $\det(A')$ along the first column, as it contains two zeros. \det(A') = 1 \cdot \det \left[ {\begin{matrix} {\sin t + 3\cos t} & {3\sin t - \cos t} \cr {3\cos t - \sin t} & {3\sin t + \cos t} \cr } } \right] - 0 + 0 Recall that the determinant of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $ad - bc$. Let $a = \sin t$ and $b = \cos t$ for simplicity. The $2 \times 2$ determinant is: $(\sin t + 3\cos t)(3\sin t + \cos t) - (3\sin t - \cos t)(3\cos t - \sin t)$ Let's calculate the two products:

• First product: $(a + 3b)(3a + b)$ $= 3a^2 + ab + 9ab + 3b^2$ $= 3a^2 + 3b^2 + 10ab$ $= 3(\sin^2 t + \cos^2 t) + 10\sin t \cos t$ Using the identity $\sin^2 t + \cos^2 t = 1$, this becomes: $= 3(1) + 10\sin t \cos t = 3 + 10\sin t \cos t$

• Second product: $(3a - b)(3b - a)$ $= 9ab - 3a^2 - 3b^2 + ab$ $= 10ab - 3a^2 - 3b^2$ $= 10\sin t \cos t - 3(\sin^2 t + \cos^2 t)$ Using the identity $\sin^2 t + \cos^2 t = 1$, this becomes: $= 10\sin t \cos t - 3(1) = 10\sin t \cos t - 3$

Now, subtract the second product from the first: $\det(A') = (3 + 10\sin t \cos t) - (10\sin t \cos t - 3)$ $\det(A') = 3 + 10\sin t \cos t - 10\sin t \cos t + 3$ $\det(A') = 6$

6. Final Determinant and Condition for Invertibility

Substituting $\det(A') = 6$ back into our expression for $\det(A)$: $\det(A) = e^{-t} \cdot 6 = 6e^{-t}$

For the matrix $A$ to be invertible, its determinant must be non-zero: $\det(A) \neq 0$ $6e^{-t} \neq 0$ We know that the exponential function $e^x$ is always positive for any real number $x$. Therefore, $e^{-t}$ is always positive ($e^{-t} > 0$) for all $t \in \mathbb{R}$. Since $e^{-t}$ is never zero, $6e^{-t}$ is also never zero for any $t \in \mathbb{R}$.

This means that the determinant of the given matrix is non-zero for all values of $t \in \mathbb{R}$. Thus, the matrix is invertible for all $t \in \mathbb{R}$.

7. Conclusion

The set of all values of $t \in \mathbb{R}$ for which the matrix is invertible is $\mathbb{R}$.

The final answer is $\boxed{\mathbb{R}}$.

Important Note for JEE Aspirants: It is crucial to meticulously perform determinant calculations, especially when trigonometric functions are involved. Even a small sign error or arithmetic mistake can lead to an incorrect result. Always double-check your algebraic expansions and trigonometric identities. In this specific problem, based on the provided matrix and standard determinant calculation, the matrix is invertible for all real values of $t$. If this result differs from a given option in an exam, it might indicate a typo in the question or options. However, always trust your thorough mathematical derivation.