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If A = [3275] and B = [-1-352], verify that (AB)–1 = B–1 A–1 - Mathematics

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प्रश्न

If A = `[(3, 2),(7, 5)]` and B = `[(-1, -3),(5, 2)]`, verify that (AB)–1 = B1 A1 

योग

उत्तर

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

|A| = 15 – 14

= 1 ≠ 0.A–1 exists.

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

A–1 = `1/|"A"|` adj A

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

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

|B| = – 2 + 15

= 13 ≠ 0.B–1 exists.

adj B = `[(2, 3),(-5, -1)]`

B–1 = `1/|"B"|` adj B

= `1/13[(2, 3),(-5, -1)]`

R.H.S : B–1A–1 = `1/13[(2, 3),(-5, -1)][(5, -2),(-7, 3)]` 

= `1/13 [(10 - 211, -4 + 9),(-25 + 7, 10 - 3)]`

= `1/13 [(-11, 5),(- 18, 7)]`  ..........(1)

L.H.S : AB = `[(3, 2),(7, 5)][(-1, -3),(5, 2)]`

= `[(-3 + 10, -9 + 4),(-7 + 25, -21 + 10)]`

= `[(7, -5),(8, -11)]`

|AB| = – 77 + 90

 = 13 ≠ 0

(AB)–1 exists.

adj AB = `[(-11, 5),(-18, 7)]`

(AB)–1 = `1/|"AB"|` adj AB

= `1/13 [(-11, 5),(- 18, 7)]`  .........(2)

(1), (2) ⇒ (AB)–1 = B–1 A–1 

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अध्याय 1: Applications of Matrices and Determinants - Exercise 1.1 [पृष्ठ १६]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 1 Applications of Matrices and Determinants
Exercise 1.1 | Q 7 | पृष्ठ १६

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