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

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

Loser66
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1@Loser66 @klimenkov gave a counterexample :p

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

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

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

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