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mprm86
Dec3-04, 10:18 PM
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.

AKG
Dec3-04, 11:06 PM
That's precisely it.

mprm86
Dec3-04, 11:13 PM
Ok, thanks.

HallsofIvy
Dec4-04, 07:22 AM
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.