Tamil Nadu Board of Secondary EducationHSC Arts Class 12

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

Sum

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

#### Solution

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

Concept: Inverse of a Non-singular Square Matrix
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Chapter 1: Applications of Matrices and Determinants - Exercise 1.1 [Page 16]

#### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 12th Mathematics Volume 1 and 2 Answers Guide
Chapter 1 Applications of Matrices and Determinants
Exercise 1.1 | Q 7 | Page 16
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