Matrix Inversion for Variation of Parameters

[gkirkland]

I am working on the following problem:

Can someone please show or explain the steps to invert the phi matrix? I've given it a few tries, but I can't reach what the book has for the answer.

Please help! Thanks

 

[SteamKing]

Phi(t)*Phi^-1(t) = I, where I is the 2x2 identity matrix.
You can use Cramer's rule to solve for the inverse by breaking the identity matrix into two columns.

If you post some of your work, we may help you spot what is going wrong with you calculations.

 

[gkirkland]

I didn't even think about using Cramer's rule to find the inverse! That was a great idea, got it on the first try.

Thanks!

 

[gkirkland]

So I calculated the inverted matrix, but for some reason I'm still not getting the correct solution. Can someone please spot my mistake?

 

[SteamKing]

I apologize for my lack of clarity in my earlier post.

In order to find the inverse of the Phi(t) matrix using Cramer's Rule, it must be used twice.

This is how the inverse is set up:
| e^-t    e^-2t|  |a11  a12|     |1  0|
|-e^-t  -2e^-2t|  |a21  a22|  =  |0  1|

where the a values represent the inverse of Phi(t). In order to calculate the a values,
the matrix equation above is split into two parts, like so:

This is how the inverse calculations are set up:
| e^-t   e^-2t|  |a11|     |1|
|-e^-t -2e^-2t|  |a21|  =  |0|
and
| e^-t   e^-2t|  |a12|     |0|
|-e^-t -2e^-2t|  |a22|  =  |1|

Cramer's Rule can be used to solve both systems. Remember, always check the calculations by multiplying Phi(t) by Phi^-1(t) to see if the Identity matrix is obtained.

I should point out that Phi^-1(t) is the inverse matrix of Phi(t) rather than 1/Phi(t)

 

[FOIWATER]

isn't the inverse of the matrix 1/det(phi) * adj(phi) not just phi?
You have to find the adjoint of phi then multiply by 1/det(phi), the adjoint matrix being the transpose of the C expansion matrix, for a 2x2 it always works out to be that you swap the signs on entry 2 and 3 and swap the placement of entry 1 and 4.

but I am not sure, trying to work it out now also not getting the books answer..