The set of all values of t `epsilon`R, for which the matrix `[(e^t,e^-t(sin t-2cost),e^-t(-2sint-cost)),(e^t,e^-t(2sin t+2cost),e^-t(sint-2cost)),(e^t,e^-t cost,e^-t sint)]` is invertible, is

Options

A. R
B. `{(2k+1)pi/2,k epsilon z}`
C. `{kpi,k epsilon z}`
D. `{kpi+pi/4,k epsilon z}`

Answer

To determine the set of all values of \( t \in \mathbb{R} \) for which the given matrix is invertible, we need to find the determinant of the matrix and see when it is non-zero.

Given the matrix:

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

### Step 1: Factor out common terms

We can factor out \( e^t \) from the first column and \( e^{-t} \) from the second and third columns:

\[ \text{det}(A) = e^t \cdot e^{-t} \cdot e^{-t} \cdot \text{det}\begin{pmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\ 1 & 2\sin t + 2\cos t & \sin t - 2\cos t \\ 1 & \cos t & \sin t \end{pmatrix} \]

This simplifies to:

\[ \text{det}(A) = e^{-t} \cdot \text{det}\begin{pmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\ 1 & 2\sin t + 2\cos t & \sin t - 2\cos t \\ 1 & \cos t & \sin t \end{pmatrix} \]

### Step 2: Row operations

We can perform row operations to simplify the determinant. Subtract the third row from the first and second rows:

\[ \text{det}(A) = e^{-t} \cdot \text{det}\begin{pmatrix} 0 & (\sin t - 2\cos t) - \sin t & (-2\sin t - \cos t) - \sin t \\ 0 & (2\sin t + 2\cos t) - \sin t & (\sin t - 2\cos t) - \sin t \\ 1 & \cos t & \sin t \end{pmatrix} \]

This simplifies to:

\[ \text{det}(A) = e^{-t} \cdot \text{det}\begin{pmatrix} 0 & -2\cos t & -3\sin t \\ 0 & \sin t + 2\cos t & 0 \\ 1 & \cos t & \sin t \end{pmatrix} \]

### Step 3: Expand the determinant

The determinant of a matrix with a row of zeros is zero. Therefore, the determinant simplifies to:

\[ \text{det}(A) = e^{-t} \cdot 0 = 0 \]

This means the determinant is zero when the first two rows are linearly dependent.

### Step 4: Condition for invertibility

For the matrix to be invertible, the determinant must be non-zero. Since \( e^{-t} \) is never zero for any real \( t \), we need to ensure that the determinant of the \( 3 \times 3 \) matrix is non-zero.

### Conclusion

The determinant will be non-zero if the rows are linearly independent. Since we have shown that the determinant simplifies to a form that can be zero, we conclude that the matrix is invertible for all \( t \in \mathbb{R} \).

Thus, the set of all values of \( t \) for which the matrix is invertible is:

\[ \text{All } t \in \mathbb{R} \]