# Help whit a problem. (I have no idea)

1. Dec 3, 2004

### mprm86

Calculate $e^A$ if $A = \left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)$

Maybe using that $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$, then

$$e^A = \sum_{n=0}^\infty \frac{\left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)^n}{n!}$$
but i dont know if this is the right way for doing this. Please help me. Thanks.

2. Dec 3, 2004

### AKG

That's precisely it.

3. Dec 3, 2004

Ok, thanks.

4. Dec 4, 2004

### HallsofIvy

Staff Emeritus
Although you may have difficulty calculating $$\left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)^n$$
Knowing that its eigenvalues are 1 and 2 (with eigenvectors <2, -1> and <1, -1> respectively) will help.