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Loser66
 one year ago
If A is an n x n matrix function such that A and \(\frac{dA}{dt}\)are the same function, then A = c\(e^tI_n\)for some constant c.
True or false? why?
Please, help
Loser66
 one year ago
If A is an n x n matrix function such that A and \(\frac{dA}{dt}\)are the same function, then A = c\(e^tI_n\)for some constant c. True or false? why? Please, help

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klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2If \(A=\left(\begin{matrix}e^{t} & 2e^{t} \\ 3e^{t}& 4e^{x}\end{matrix}\right)\), what is \(\frac{dA}{dt}\)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{dA}{dt}=A ~~but~~A\neq ce^tI_n\]therefore, it's false, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2the area of ACDE = area ABE + area BCDE = AB *EB/2 + EB*BC = 4*4\(\frac{\sqrt{3}}{2}\)+4\(\sqrt{3}\)*11

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2since AB=4 , BC=11. got it? it is = 8\(\sqrt{3}\)+44\(\sqrt{3}\)=52\(\sqrt{3}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2@MayMay_69 that's yours. Take it, I am gonna delete it.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2@oldrin.bataku am I right for my own problem?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2I am sorry for messing up. May may asked for help but he/she has no link, I used my post to anser him/her

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1@Loser66 @klimenkov gave a counterexample :p

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2that means the answer is "False" right?

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1it need not be \(A=ce^tI_n\) merely \(A=e^tB\) for constant matrix \(B\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2thank you very much. @oldrin.bataku

MayMay_69
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you very very much loser

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2You got what I mean, right? @MayMay_69
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