Calculate the inverse of [1, i/2 ; -1/3, 1] = I - [0, -i/2 ; 1/3, 0] using a power series.
I know that (I - A)^-1 = I + A + A^2 +A^3 but am unsure of how to calculate this without using a computer.
Thanks
Calculate the inverse of [1, i/2 ; -1/3, 1] = I - [0, -i/2 ; 1/3, 0] using a power series.
I know that (I - A)^-1 = I + A + A^2 +A^3 but am unsure of how to calculate this without using a computer.
Thanks
There are a few ways to do this. Typically you want to diagonalize any matrix that you're going to raising to powers, but here you can avoid that.
$\displaystyle \text{Let } A = \left(\begin{matrix}0 & -i/2 \\ 1/3 & 0 \end{matrix}\right). \text{ Then } A^2 = cI, \text{ where } c = \frac{-i}{6}.$
$\displaystyle \text{Thus }\left(\begin{matrix}1 & i/2 \\ -1/3 & 1 \end{matrix}\right)^{-1}$
$\displaystyle = (I - A )^{-1}$
$\displaystyle = I + A + A^2 + A^3 +...$
Spoiler:
I just did the matrix multiplication (they're 2x2 matricies with half zeros, so that's about than 10 second of work). I didn't know ahead of time it would work out this way. If was only after I did the multiplication that I saw that the power series would thereby algebraically "collapse" the way it did.
As an aside, that's a kinda universal approach I have to math problems. When I'm given something general or abstract, I try to do (or apply it to) a few specific, more concrete examples - or in this case compute a term or two of a series. That often helps me "see" what's going on. Sometimes, like with this problem, I'll find something unexpected that will give me the key to how to solve the problem.