mprm86
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Calculate [itex]e^A[/itex] if [itex]A = \left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)[/itex]
Maybe using that [tex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex], then
[tex]e^A = \sum_{n=0}^\infty \frac{\left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)^n}{n!}[/tex]
but i don't know if this is the right way for doing this. Please help me. Thanks.
Maybe using that [tex]e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex], then
[tex]e^A = \sum_{n=0}^\infty \frac{\left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)^n}{n!}[/tex]
but i don't know if this is the right way for doing this. Please help me. Thanks.