Inverse of matrix with varying parameters

Question

Ok so I need some sort of verification on this. I have run into this matrix $\begin{pmatrix} e^t&3e^{-t}\\e^t&e^{-t} \end{pmatrix}$ and I need to find the inverse of this matrix. The book simply ignores the fact that there is a function in the matrix and computes the inverse normally (without the functions). However when we start throwing in $\sin t$ and $\cos t$ we start to see complicated inverses. What exactly is the procedure to determine the inverse of these types of matrices and how do you know if it is "safe" to take the inverse of the fundamental matrix? These matrices come up in differential equations for solutions to non-homogenous systems. How would you take the inverse had it been $\begin{pmatrix} \sin t& 3\cos t \\ \sin t& \cos t\end{pmatrix}$ ?

Answer 1

Hint

Do you know that

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-c\\-b&a\end{pmatrix},\qquad \text{if}\;\; ad-bc\ne0$$

Answer 2

To make a long story short, everything you learned before this point with numbers instead of functions is more or less the same. In this case, we use that fact that a matrix can be inverted if it has none zero determinate. So the standard inversion matrix formula holds.

$det = ad - bc = e^t e^{-t} - 3e^t e^{-t} = e^{0} - 3 e^{0} = -2 \neq 0 \quad \forall t \in \mathbb{R}$