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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![a = [ cc 3&1 -1&2 ]](https://static.philoid.co/ncertusercontent/solutions/?domain=gF&l=PROJ28781/1554209799326247.png)
![a = [ ll 3&1 1&2 ]](https://static.philoid.co/ncertusercontent/solutions/?domain=gF&l=PROJ28781/1554209802337249.png)