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:

We have the square matrix

$$A=\begin{bmatrix} e^{t} & e^{-t}\cos t & e^{-t}\sin t\\[4pt] e^{t} & -e^{-t}\cos t-e^{-t}\sin t & -e^{-t}\sin t+e^{-t}\cos t\\[4pt] e^{t} & 2e^{-t}\sin t & -2e^{-t}\cos t \end{bmatrix}.$$

A matrix is invertible exactly when its determinant is non-zero. Hence we compute $$\det(A).$$

First, in every column we take out the common exponential factor:

Column 1 has a common factor $$e^{t},$$ Column 2 has a common factor $$e^{-t},$$ Column 3 has a common factor $$e^{-t}.$$

The rule “pulling a scalar $$k$$ out of a column multiplies the determinant by $$k$$” gives

$$\det(A)=e^{t}\,e^{-t}\,e^{-t}\;\det\!\begin{bmatrix} 1 & \cos t & \sin t\\[4pt] 1 & -\cos t-\sin t & -\sin t+\cos t\\[4pt] 1 & 2\sin t & -2\cos t \end{bmatrix}.$$

Simplifying the scalar product, $$e^{t}e^{-t}e^{-t}=e^{-t},$$ so

$$ \det(A)=e^{-t}\,\det(B), $$ where

$$B=\begin{bmatrix} 1 & \cos t & \sin t\\[4pt] 1 & -\cos t-\sin t & -\sin t+\cos t\\[4pt] 1 & 2\sin t & -2\cos t \end{bmatrix}.$$

Because $$e^{-t}\neq 0$$ for every real $$t,$$ the sign of $$\det(A)$$ is the same as that of $$\det(B).$$ Now we find $$\det(B).$$

Subtracting the first row from the second and third (a row operation that does not change the determinant) gives

$$\det(B)=\det\!\begin{bmatrix} 1 & \cos t & \sin t\\[4pt] 0 & (-\cos t-\sin t)-\cos t & (-\sin t+\cos t)-\sin t\\[4pt] 0 & 2\sin t-\cos t & -2\cos t-\sin t \end{bmatrix}.$$

Thus

$$B=\begin{bmatrix} 1 & \cos t & \sin t\\[4pt] 0 & -2\cos t-\sin t & \cos t-2\sin t\\[4pt] 0 & 2\sin t-\cos t & -2\cos t-\sin t \end{bmatrix}.$$

Expanding the determinant down the first column (using the formula “minor times cofactor,” and noting that the sign is $$+$$ because the element is in position (1,1)):

$$\det(B)=1\cdot \det\!\begin{bmatrix} -2\cos t-\sin t & \cos t-2\sin t\\[4pt] 2\sin t-\cos t & -2\cos t-\sin t \end{bmatrix}.$$

For the 2 × 2 determinant we use $$\det\!\begin{bmatrix}p&q\\r&s\end{bmatrix}=ps-qr:$$

$$\det(B)=(-2\cos t-\sin t)(-2\cos t-\sin t)\,- (\cos t-2\sin t)(2\sin t-\cos t).$$

Introduce the abbreviations

$$ a=-2\cos t-\sin t,\qquad b=\cos t-2\sin t. $$

Notice that $$2\sin t-\cos t=-b.$$ Therefore

$$ \det(B)=a^{2}-b(-b)=a^{2}+b^{2}. $$

The expression $$a^{2}+b^{2}$$ is a sum of two squares, which is never negative, and it equals zero only if both $$a=0$$ and $$b=0.$

Setting $$a=0$$ gives $$-2\cos t-\sin t=0\;\Longrightarrow\;\tan t=-2.$$ Setting $$b=0$$ gives $$\cos t-2\sin t=0\;\Longrightarrow\;\tan t=\dfrac12.$$

Because $$\tan t$$ cannot be simultaneously $$-2$$ and $$\dfrac12,$$ the pair of equations has no common solution. Hence $$a^{2}+b^{2}>0$$ for all real $$t,$$ so

$$ \det(B)\;>\;0\quad\text{for every real }t. $$

Finally, since $$\det(A)=e^{-t}\det(B)$$ and neither factor is ever zero, we have

$$ \det(A)\neq 0\quad\text{for every real }t. $$

Thus the matrix $$A$$ is invertible for all real values of $$t.$$

Hence, the correct answer is Option D.