Question

Let A = `[(3,7),(2,5)]` and B = `[(6,8),(7,9)]`. Verify that (AB)⁻¹ = B⁻¹A⁻¹.

Answer 1

A = `[(3,7),(2,5)]`

|A| = 15 − 14

= 1 ≠ 0

∴ A⁻¹ exists.

A⁻¹ = `1/|A|` (adj A)

= `1/1 [(5,-7),(-2,3)]`

= `[(5,-7),(-2,3)]`

B = `[(6,8),(7,9)]`

|B| = 54 − 56

= −2 ≠ 0

∴ B⁻¹ exists.

B⁻¹ = `1/|B|` (adj B)

= `1/-2 [(9,-8),(-7,6)]`

= `[(-9/2,4),(7/2, -3)]`

B⁻¹A⁻¹ = `[(-9/2,4),(7/2, -3)][(5,-7),(-2,3)]`

= `[(-45/2 - 8, 63/2 + 12),(35/2 + 6, -49/2 - 9)]`

= `[(-61/2, 87/2),(47/2, -67/2)]`

AB = `[(3,7),(2,5)][(6,8),(7,9)]`

= `[(18 + 49, 24 + 63),(12 + 35, 16 + 45)]`

= `[(67,87),(47,61)]`

|AB| = 4087 − 4089

= −2 ≠ 0

∴ (AB)⁻¹ exists.

(AB)⁻¹ = `1/|(AB)|` (adj AB)

= `1/-2 [(61,-87),(-47,67)]`

`= [(-61/2,87/2),(47/2,-67/2)]`

Hence, (AB)⁻¹ = B⁻¹A⁻¹