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

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प्रश्न

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

योग
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उत्तर

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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अध्याय 4: Determinants - Exercise 4.5 [पृष्ठ १३२]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 4 Determinants
Exercise 4.5 | Q 12 | पृष्ठ १३२

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