Find the inverse of the matrices (if it exists). [(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)] - Mathematics

Question

Find the inverse of the matrices (if it exists).

`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`

Sum

Solution

Let A = `[(1, 0, 0), (0, cos alpha, sin alpha), (0,sin alpha, -cos alpha)]`

|A| = 1(−cos² α − sin² α)

= −1 ≠ 0

∴ A⁻¹ exists.

C_ij = (−1)^{i + j} M_ij

C₁₁ = `(-1)^(1 + 1) |(cos alpha, sin alpha), (sin alpha, -cos alpha)|`

= −cos²α − sin ²α

= −1

C₁₂ = `(-1)^(1 + 2) |(0, sin alpha), (0, -cos alpha)|` = 0

C₁₃ = `(-1)^(1 + 3) |(0, cos alpha), (0, sin alpha)|` = 0

C₂₁ = `(-1)^(2 + 1) |(0, 0), (sin alpha, -cos alpha)|` = 0

C₂₂ = `(-1)^(2 + 2) |(1, 0), (0, -cos alpha)|` = −cos α

C₂₃ = `(-1)^(2 + 3) |(1, 0), (0, sin alpha)|` = −sin α

C₃₁ = `(-1)^(3 + 1) |(0, 0), (cos alpha, sin alpha)|` = 0

C₃₂ = `(-1)^(3 + 2) |(1, 0), (0, sin alpha)|` = −sin α

C₃₃ = `(-1)^(3 + 3) |(1, 0), (0, cos alpha)|` = cos α

Now, Adj A = `[(-1, 0, 0), (0, -cos alpha, -sin alpha), (0, -sin alpha, cos alpha)]`

A⁻¹ = `1/|A|` (adj A)

= `1/-1 [(-1, 0, 0), (0, -cos alpha, -sin alpha), (0, -sin alpha, cos alpha)]`

= `[(1, 0, 0), (0, cos alpha, sin alpha), (0, sin alpha, -cos alpha)]`

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Properties of Matrix Multiplication - Inverse of a Square Matrix by the Adjoint Method

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Q 11 Q 10 Q 12

Chapter 4: Determinants - Exercise 4.5 [Page 132]

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NCERT Mathematics Part 1 and 2 [English] Class 12

Chapter 4 Determinants
Exercise 4.5 | Q 11 | Page 132

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