mrbassman said:
i wish it was
| e^0 e^-30 |
| |
| e^0 e^-1/15 | like i thought it was
Don't we all
There's a couple of ways of doing this:
1. Diagonalising A into the form A = T*D*T^-1, where T is the matrix whose columns are eigenvectors of A, and D is a diagonal matrix whose entries are eigenvalues of A. Then exp(At) = exp(T*(Dt)*T^-1) = T*exp(Dt)*T^-1. exp(Dt) is easy to work out since D is a diagonal matrix, so it's just equal to the exponential of each of the diagonal elements inside (this method FAILS when the matrix A has repeated eigenvalues since T is not invertible, so in those cases you need to use Jordan forms which are a pain in the arse)
2. Using Laplace Transforms: Laplace{e^(At)} = (sI - A)^-1, or e^(At) = Inverse Laplace{(sI - A)^-1}
This method is somewhat easier (less messy) to use and is ALWAYS applicable (unlike the eigenvalues method), but you need to know a bit about Laplace Transforms. I prefer this method.
Using method 2,
A = [0, -30; 0, 1/15]
(sI - A) = [s, 0; 0, s] - [0, -30; 0, 1/15]
= [s, 30; 0, s - 1/15]
inverting (sI - A),
(sI-A)^-1 = 1/(s^2 - s/15) * [s - 1/15, -30; 0, s]
= 1/(s*(s - 1/15)) * [s - 1/15, -30; 0, s]
you have to now decompose this into partial fractions, using the "cover up" rule:
= -15 * [-1/15, -30; 0, 0] / s + 15 * [0, -30; 0, 1/15] / (s - 1/15)
= [1, 450; 0, 0] / s + [0, -450; 0, 1] / (s - 1/15)
Now you have to use Inverse Laplace (**):
e^(At) = Inverse Laplace{[1, 450; 0, 0] / s + [0, -450; 0, 1] / (s - 1/15)}
= [1, 450; 0, 0] + [0, -450; 0, 1] * e^(t/15)
which is the answer hopefully...
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** For a general function of t, Laplace transform is defined as:
Laplace{f(t)} = Integral{0 -> Infinity} f(t)*e^(-st) dt
The Laplace transform of a matrix is the Laplace transform of each element in the matrix.
For determining this matrix exponential, all we used was a certain special case: (prove it for yourself)
Laplace{e^at} = 1/(s - a)