Let . Verify that (AB)–1 = B–1 A–1.
We have AB = = (61)(67)-(47)(87) = -2
Here determinant of matrix = |AB|≠ 0 hence (AB)-1 exists.
Also |A| = 1 ≠ 0 and |B| = -2 ≠ 0.
∴ A-1 and B-1 will also exist and are given by-
And hence,
{Hence proved}