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Find A–1 if A = [011101110] and show that A–1 = AIA2-3I2.

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Question

Find A–1 if A = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)]` and show that A–1 = `("A"^2 - 3"I")/2`.

Sum
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Solution

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

Co-factors are:

A11 = –1,

A12 = 1

A13 = 1

A21 = 1

A22 = –1

A23 = 1

A31 = 1

A31 = 1

A32 = 1

A33 = –1

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

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

|A| = 0 – 1(–1) + 1.1 = 2

∴ A–1 = `("adj A")/|"A"|`

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

Now, A2 = `[(0, 1, 1),(1, 0, 1),(1, 1, 0)] * [(0, 1, 1),(1, 0, 1),(1, 1, 0)]`

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

∴ `("a"^2 - 3"I")/2 = 1/2{[(2, 1, 1),(1, 2, 1),(1, 1, 2)] - [(3, 0, 0),(0, 3, 0),(0, 0, 3)]}`

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

= A–1

Hence proved.

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Chapter 4: Determinants - Exercise [Page 79]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 4 Determinants
Exercise | Q 17 | Page 79

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