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}


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