# How can i calculate e^A*t

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Nick on 30 May 2012
Commented: Walter Roberson on 12 Jun 2022
How can i calculate e^A*t without using Markov Chain?
Where e=exp , A is a square matrix, and t is a variable
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Walter Roberson on 12 Jun 2022
Please expand on your question ?

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### Accepted Answer

Elad on 30 May 2012
exp(A.*t)
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### More Answers (5)

Kye Taylor on 31 May 2012
Use the expm function for computing a matrix exponential
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Walter Roberson on 9 Nov 2017
We tried a number of times to get the original poster to clarify, but all we got was that they want the exp() solution and that they are looking for a "deeper reason" for something. The poster effectively defined the exp() solution as being the correct one.
Your analysis might well be what the poster really needed, but it is contrary to what little they defined as being correct for their needs.

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Junsheng SU on 28 Nov 2017
syms t; expm(A*t);
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Lars Nagel on 17 Aug 2019
Worked perfectly for me!

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Shenhai on 20 Jan 2017
Edited: Shenhai on 20 Jan 2017
I guess it is not always possible to get the close form solution of exp(At)...
Sometimes I can get result with: exp(At) = iL(sI-A)^-1, where iL is the inverse Laplace transformation, like:
syms s t
A = [0 1;0 0];
expAt = ilaplace(inv(s*eye(size(A,1))-A),s,t);
This will give the result as: [1 t;0 1]
Any other ideas?
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Shahram Bekhrad on 8 Jun 2012
As far as I'm aware you probably need it for finding the answer of a state space equation. I myself couldn't find any good function or command yet, so you might have to write a Script file (m-file) and find it. you can use about 3 or 4 way of calculating the said statement. These things are taught in courses like modern control theory. I used the following expression but still have some difficulties. exp(A.t)=I+At+ (At)^2/2! + (At)^3/3!+ (At)^4/4!+. . .
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ABCD on 29 Sep 2016
Dear Nick, do you mean this?
>> a = [1 2 3 ; 2 5 2; 1 4 3]
a =
1 2 3
2 5 2
1 4 3
>> syms t >> exp(a*t)
ans =
[ exp(t), exp(2*t), exp(3*t)] [ exp(2*t), exp(5*t), exp(2*t)] [ exp(t), exp(4*t), exp(3*t)]
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ABCD on 29 Sep 2016
>> a = [1 2 3 ; 2 5 2; 1 4 3]
a =
1 2 3
2 5 2
1 4 3
>> syms t
>> exp(a*t)
ans =
[ exp(t), exp(2*t), exp(3*t)]
[ exp(2*t), exp(5*t), exp(2*t)]
[ exp(t), exp(4*t), exp(3*t)]

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