Calculate $e^{\mathbb{A}}$ where ${\mathbb{A}}$ is the matrix of the following differential system

Question

we have the following system of differential equations :

$\begin{pmatrix} x\\ y \end{pmatrix} '$= $\begin{pmatrix} 3 &-2 \\ 2& -2 \end{pmatrix}$$\begin{pmatrix} x\\ y \end{pmatrix}$

Let $\mathbb{A}$ be the matrix of this system , I have to calculate $\exp(\mathbb{A})$

Here is what I know :

$\exp(\mathbb{A}) = \sum{\frac{\mathbb{A}^{n}}{n!}}$

Example for the identity :

$\mathbb{A}=\begin{pmatrix} 1 &0 \\ 0& 1 \end{pmatrix}\;\;$,$\; \exp(\mathbb{A})=I + \mathbb{A} +\frac{\mathbb{A}}{2} + ...... $

So :

$\exp^{\begin{pmatrix} 1 &0 \\ 0& 1 \end{pmatrix}}= \begin{pmatrix} 1+\frac{1}{1!}+\frac{1}{2!}+.....&0 \\ 0& 1+\frac{1}{1!}+\frac{1}{2!}+..... \end{pmatrix} $ = $\begin{pmatrix} e &0 \\ 0& e \end{pmatrix}$

I tried the same thing for my problem but it didn't work , any help would be a lot appreciated. Thanks in advance

Answer 1

HINT

If you can write $A = VDV^{-1}$ (diagonalization), then $$ A^2 = A\cdot A = \left(VDV^{-1}\right)\left(VDV^{-1}\right) = VD\left(V^{-1}V\right)DV^{-1} = VD^2V^{-1} $$ and similarly $A^n = VD^nV^{-1}$, so you have $$ e^A = \sum_{k=0}^\infty \frac{A^k}{k!} = \sum_{k=0}^\infty \frac{VD^kV^{-1}}{k!} = V \left( \sum_{k=0}^\infty \frac{D^k}{k!} \right) V^{-1} = V e^{D} V^{-1}. $$

Now can you diagonalize $A$, find $D$ and $V$, and plug in?