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