Tamil Nadu Board of Secondary EducationHSC Arts Class 12

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

Sum

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

#### 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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