Tamil Nadu Board of Secondary EducationHSC Arts Class 12

If A = [53-1-2], show that A2 – 3A – 7I2 = O2. Hence find A–1 - Mathematics

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Sum

If A = `[(5, 3),(-1, -2)]`, show that A2 – 3A – 7I2 = O2. Hence find A–1 

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Solution

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

A2 = A × A

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

= `[(25 - 3, 15 - 6),(-5 + 2, -3 + 4)]`

= `[(22, 9),(-3, 1)]`

 – 3A = `-3[(5, 3),(-1, -2)]`

= `[(-15, -9),(3, 6)]`

– 7I2 = `-7[(1, 0),(0, 1)]`

= `[(-7, 0),(0, -7)]`

A2 – 3A – 7I2 

= `[(22, 9),(-3, 1)] + [(15, -9),(3, 6)] + [(-7, 0),(0, -7)]`

= `[(0, 0),(0, 0)]`

∴ A2 – 3A – 7I2 = O2

Post multiply this equation by A–1

A2A– 3A A– 7IA1 = 0

A – 3I – 7A1 = 0

A – 3I = 7 A

A1 = `1/7 ("A" - 3"I")`

= `1/7 [((5, 3),(-1, -2)), -3((1, 0),(0, 1))]`

= `1/7 [((5, 3),(-1, -2)) + ((-3, 0),(0, -3))]`

A1 = `1/7[(2, 3),(-1, -5)]`

Concept: Inverse of a Non-singular Square Matrix
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Chapter 1: Applications of Matrices and Determinants - Exercise 1.1 [Page 15]

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 4 | Page 15

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