Tamil Nadu Board of Secondary EducationHSC Arts Class 12

If A = [011101110], show that AAIA-1=12(A2-3I) - Mathematics

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Sum

If A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]`, show that `"A"^-1 = 1/2("A"^2 - 3"I")`

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Solution

A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]`

A2 = A × A

= `[(0, 1, 1),(1, 0, 1),(1, 1, 0)] [(0, 1, 1),(1, 0, 1),(1, 1, 0)]`

= `[(0 + 1 + 1, 0 + 0 + 1, 0 + 1 + 0),(0 + 0 + 1, 1 + 0 + 1, 1 + 0 + 0),(0 + 1 + 0, 1 + 0 + 0, 1 + 1 + 0)]`

= `[(2, 1, 1),(1, 2, 1),(1, 1, 2)]`

A2 – 3I = `[(2, 1, 1),(1, 2, 1),(1, 1, 2)] - 3[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

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

A2 – 3I = `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`  .........(1)

adj A = `[(+|(0, 1),(1, 0)|, -|(1, 1),(1, 0)|, +|(1, 0),(1, 1)|),(-|(1, 1),(1, 0)|, +|(0, 1),(1, 0)|, -|(0, 1),(1, 1)|),(+|(1, 1),(0, 1)|, -|(0, 1),(1, 1)|, +|(0, 1),(1, 0)|)]^"T"`

= `[(+(0 - 1) , -(0 - 1), +(1 - 0)),(-(0 - 1), +(0 - 1), -(0 - 1)),(+(1 - 0), -(0 - 1), +(0 - 1))]^"T"`

= `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]^"T"`

adj A = `[(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`

A–1 = `1/|"A"|` adj A = `1/2 [(-1, 1, 1),(1, -1, 1),(1, 1, -1)]`

A–1 = `1/2 ("A"^2 - 3"I")`  ......(Using (1))

Hence proved

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 14 | Page 16

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