Find $\exp(D)$ where $D = \begin{bmatrix}5& -6 \\ 3 & -4\end{bmatrix}. $ The question is 

Find $\exp(D)$ where $D = \begin{bmatrix}5& -6 \\ 3 & -4\end{bmatrix}. $

I am wondering does finding the $\exp(D)$ requires looking for the canonical form... Could someone please help?

Hint:


*

*Write the Jordan Normal Form (it is diagonalizable) with unique eigenvalues.

*$e^{D} = P \cdot e^J \cdot P^{-1}$


The Jordan Normal Form is:
$$A = P J P^{-1} = \begin{bmatrix}1&2 \\ 1 & 1\end{bmatrix}~\begin{bmatrix}-1&0 \\0 & 2\end{bmatrix}~\begin{bmatrix}-1&2 \\ 1 & -1\end{bmatrix}$$
Now use the above formula to find the exponential of $D$.
