For 2×2 matrix, A=[a b, c d]

we want to find B such that AB = I

Let B = [e f, g h]

then ae+bg=1

af+bh=0

ce+dg=0

cf+dh=1

from (2) and (3), we have

f=-bh/a and e=-dg/c

Put e into the (1),

-adg/c + bg = 1

-adg + bcg = c

g(bc-ad) = c

g = c/(bc-ad)

And we find a repeating occurrence of ‘bc-ad’, named as the determinant.

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