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

1 Answer
George C.
Jun 9, 2018

#((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))))#

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