In case you didn't know . . .
Find the inverse of: .
 Switch the two on the main diagonal
 Change the signs of the other two: .
 Divide everything by the determinant: .
being somewhat bad at "memorizing things" i often have to "re-derive" this formula. here is how i do it:
suppose A =
i want to find B =
[z w], so that AB = I. so:
ax + bz = 1
ay + bw = 0
cx + dz = 0
cy + dw = 1
that's 4 linear equations in 4 unknowns (x,y,z,w). if it has a solution, there will be only one.
from the first equation, and the third, we have:
adx + bdz = d
-bcx - bdz = 0
thus (ad - bc)x = d. if ad - bc ≠ 0, this gives:
x = d/(ad - bc).
from the second, and fourth equation, we get:
ady + bdw = 0
-bcy - bdw = -b
thus (ad - bc)y = -b, and again if ad - bc ≠ 0, this gives us:
y = -b/(ad - bc).
to solve for z, we use equations one and three again:
acx + bcz = c
-acx - adz = 0
thus (bc - ad)z = c, so
z = c/(bd - ac) = -c/(ad - bc)
finally, from equations two and four:
acy + bcw = 0
-acy - adw = -a
(bc - ad)w = -a
w = -a/(bc - ad) = a/(ad - bc).
it should come as no surprise that is is 1/(det(A) (= 1/(ad - bc)) times:
[-c a], as Soroban points out above.