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

Question

3956 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T00:28:30

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

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:

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

How do I find the inverse of a 2xx22xx2 matrix?