Question

Can you check my answer?

Calculate \(e^{tA}\) with A={3,-2;2,-2} by the method of your choice.

Answer 1

I got \[e^{tA}=\left[\begin{matrix}{\frac{-1}{3}e^{2t}+\frac{4}{3}e^{-t}} & {e^{2t}-e^{-t}} \\ {-e^{2t}+e^{-t}} & {\frac{4}{3}e^{2t}-\frac{1}{3}e^{-t}}\end{matrix}\right]\]

Answer 2

@ganeshie8 , do you mind just a quick check?

Answer 3

its my fav subject :3

Answer 4

good to know

Answer 5

matrix exponentials! @dan815 still rememeber these from LA ?

Answer 6

haha

Answer 7

are there any initial conditions ?

Answer 8

nope that''s it. I'm scanning my work now

Answer 9

ok , i''ll check for u :D

Answer 10

here is my work

Answer 11

wolfram agrees with your answer
http://gyazo.com/5cabd46e2faec6d3336bf6fc4bb405ef

Answer 12

But I don't have any of the c's

Answer 13

*more or less it matches upto a constant, but i found this solution by solving the system without initial conditions

Answer 14

but there are no initial conditions

Answer 15

let me pull up my LA notes

Answer 16

But my notes don't have a c either

Answer 17

oh u used Laplace inverse ,using psi function is much easy

Answer 18

we didn't learn that

Answer 19

(I don't think so at least)

Answer 20

kk , its shortcut way for ur solution :)
since u dint take yet u wont understand it until u've prove things first xD
so leave it, ur solution looks good to me :|

Answer 21

awesome, thank you. Out of curiosity, what is the psi way?

Answer 22

its a function been used as solution to solve some cases of ODE , homogeneous and non

Answer 23

to get e^tA , we only need to find eigen values and vectores with some trick from psi

Answer 24

darn, that would have been easier