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.
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.