# Inverting a matrix

• Oct 13th 2008, 01:00 AM
horseypie
Inverting a matrix
a 1
1 -a

i find the inverse, which is

a/(a^2 + 1) 1/(a^2 + 1)
1/(a^2 + 1) -a/(a^2 + 1)

which is correct, no?

so can i multiply that matrix by $(a^2 + 1)$ and get this matrix

a 1
1 -a

which is the same as the initial matrix? so is it true that that matrix is its own inverse?

Sorry guys but i cant seem to work out how to do matrices in here properly
• Oct 13th 2008, 01:25 AM
Cloud_pin
Just checking I'm following what you're after: if you call your original matrix A, and your inverse is A^-1, you want A^-1A = I?
If so, your inverse is

-a/(-a^2 - 1) -1/(-a^2 - 1)
-1/(-a^2 - 1) a/(-a^2 - 1)

If your original matrix was its own inverse, then AA=I, but in your case AA =
a^2 + 1 0
0 1 + a^2
Which is (a^2 + 1)I, but not I itself.
• Oct 13th 2008, 01:28 AM
horseypie
yeah thats right...thanks heaps mate

i think ive worked out what i need to know now, thanks heaps