Consider a matrix A(θ) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mi>sin</mi><mi>θ</mi></mtd><mtd><mi>cos</mi><mi>θ</mi></mtd></mtr><mtr><mtd><mo>-</mo><mi>cos</mi><mi>θ</mi></mtd><mtd><mi>sin</mi><mi>θ</mi></mtd></mtr></mtable></mfenced></math> then
(A) A(θ) is symmetric
(B) A(θ) is skew symmetric
(C) A<sup>-1</sup>(θ) = A(π - θ)
(D) A<sup>2</sup>(θ) = A(π/2 - 2θ)
Step-by-Step Solution
Key Concept: Check properties of the given matrix A(\theta). A(\theta) is not symmetric or skew-symmetric. Calculate A⁻^1(\theta) and compare with A(\pi-\theta).
A(\theta) = [[sin \theta, cos \theta], [-cos \theta, sin \theta]]. Aᵀ(\theta) = [[sin \theta, -cos \theta], [cos \theta, sin \theta]]. Since Aᵀ(\theta) \neq A(\theta) and Aᵀ(\theta) \neq -A(\theta), it is neither symmetric nor skew-symmetric. det(A(\theta)) = sin^2\theta + cos^2\theta = 1. A⁻^1(\theta) = (1/det(A)) * adj(A) = [[sin \theta, -cos \theta], [cos \theta, sin \theta]]. Now, A(\pi - \theta) = [[sin(\pi - \theta), cos(\pi - \theta)], [-cos(\pi - \theta), sin(\pi - \theta)]] = [[sin \theta, -cos \theta], [cos \theta, sin \theta]]. Thus, A⁻^1(\theta) = A(\pi - \theta).
Correct Answer: 3