Question

How do I find the inverse of a `2xx2` matrix?

Answer 1

`((a,b),(c,d))^(-1) = 1/abs((a,b),(c,d)) ((d, -b), (-c, a))`

Explanation:

Let's have a go a this without simply plugging in a remembered formula.

Given a matrix `((a, b), (c, d))`, let's try multiplying it by `((d, -b),(-c,a))` and see what we get:

`((a, b),(c, d))((d, -b),(-c, a)) = ((ad-bc, 0), (0, ad-bc))`

So if we multiply by `1/(ad-bc) = 1/abs((a,b),(c,d))` then we find the inverse matrix:

`1/abs((a,b),(c,d)) ((d, -b),(-c, a)) = ((d/abs((a,b),(c,d)), -b/abs((a,b),(c,d))), (-c/abs((a,b),(c,d)), a/abs((a,b),(c,d))))`