mprm86
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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 don't 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 don't know if this is the right way for doing this. Please help me. Thanks.