Help whit a problem. (I have no idea)

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Discussion Overview

The discussion revolves around calculating the matrix exponential \( e^A \) for the matrix \( A = \left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right) \). Participants explore methods for this calculation, including the series expansion and the use of eigenvalues and eigenvectors.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using the series expansion \( e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \) to calculate \( e^A \).
  • Another participant confirms that the proposed method is correct.
  • A later reply indicates that calculating \( \left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)^n \) may be challenging, but knowing the eigenvalues (1 and 2) and their corresponding eigenvectors could assist in the process.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the series expansion method, but there is no consensus on the ease of calculating the powers of the matrix or the best approach to take.

Contextual Notes

Participants mention the eigenvalues and eigenvectors, but do not provide detailed calculations or confirm the implications of these values on the matrix exponential.

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.
 
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That's precisely it.
 
Ok, thanks.
 
Although you may have difficulty calculating [tex]\left(\begin{array}{cc}0 & -2\\1 & 3\end{array}\right)^n[/tex]
Knowing that its eigenvalues are 1 and 2 (with eigenvectors <2, -1> and <1, -1> respectively) will help.
 

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