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

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

• Do you know how to diagonalize a matrix? Commented Feb 27, 2020 at 20:59
• hint: diagonalize $A$. Commented Feb 27, 2020 at 20:59
• @J.W.Tanner Yes we've done it to find the solution of a differential equation using matrix .. my bad
– user730480
Commented Feb 27, 2020 at 21:09
• The next topic is to find the exponential function of a matrix that is not diagonalizable. Commented Feb 27, 2020 at 22:04

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?