That’s right, time to help the irresponsible student with her exam prep! Actually, I’m not sure this will even be on the exam. But it’s bothering me and I can’t concentrate on the important stuff.
Suppose we have these matrices. You are given Matrix A and Matrix B, and you are asked to find C if AC = B. It’s fairly simple to solve for 2x2 matrices, but how do you do 3x3s? My wonderfully helpful textbook tells me to just leave it up to my calculator, but the calculators we have can’t handle that sort of thing (they can do matrix multiplication but they can’t find inverse matrices). I seem to recall some sort of method involving the identity matrix but I can’t. Freaking. Remember. Argh.
I’m not sure how you’d do it without finding the inverse of A. I suppose you know that, if AC = B, then C = A[sup]-1[/sup]B.
To find the inverse of A, you can set up a 3x6 matrix with A on the left and I (the identity matrix) on the right, and do the row operations required to turn the left side into I (a la the Gauss-Jordan elimination method), and what shows up on the right will be A[sup]-1[/sup].
Are you sure your calculator won’t find inverses? I would have thought that any calculator that would do matrix multiplication could also find inverses.
Actually, I guess if you really wanted to do it without finding an inverse matrix, you could write C with variables standing for all nine entries, work out what AC would be, set its entries equal to those of B, and solve the resulting systems of equations. But that would be at least as much work as finding A[sup]-1[/sup].
Well then. I don’t think whatever method we were taught worked for all 3x3 matrices, it was just a shortcut for certain kinds… man, it probably doesn’t help any that I haven’t looked at matrices since March, does it? Sigh. Thank Og these are my mock finals.
The method outlined by Thudlow Bink works well for all square matrices. It’s very simple to compute the inverse of a matrix with this method, too. Suppose you had a 2x2 matrix like so:
a b
c d
Simply add an identity matrix to the right hand side, so the old matrix becomes:
a b 1 0
c d 0 1
and, using elementary row operations, get the matrix so that it looks like so:
1 0 e f
0 1 g h
where e f g and h are the entries of your inverse matrix. It’s the same principle for 3x3 matrices, too. You can now use your inverse to compute the matrix C.
No matter what you do, you’re essentially inverting a matrix to do it – you’ll do the same math, even if it’s not obvious at the time that this is true.
Google “matrix inversion” and you get some sites on it, even ones like this that give explicit algebraic formulae:
Ohh, thanks for the help, everyone I did look at that Wikipedia page and to tell the truth, it hurt my brain. I’m not the most mathematically-minded person and abstract stuff trips me up easily. I still swear we had a faster way of working out solutions by hand but I don’t suppose it matters anymore since my friend has just pointed out that our calculators can indeed do inverses and I have somehow managed to get through a whole year (and several tests where this knowledge would have come in handy) without discovering this. Now if you’ll excuse me, I’m off to beat my head against a wall.
Here’s another formula that works to calculate the inverse. It’s more computation intensive than the elementary row operations method, but you never have to guess what move to make next[1].
For each entry a[sub]ij[/sub] in the matrix, strike out the ith row and jth column. This leaves an n-1 by n-1 matrix. Take its determinant and call it A[sub]ji[/sub]. Yes, that’s not a typo in the subscript – put it in the jth column and ith row of a new matrix.
Now take the determinant of the whole original matrix and call it A. Divide A into each entry of the new matrix. The resulting matrix is the inverse of the original matrix.
[1] Truth be told, the other method you don’t have to guess if you know what you’re doing, but it can be annoying if you’re unused to it.