<p>If \(A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}\), then match Column I with Column II:</p><p><b>Column I:</b><br>(A) adj A<br>(B) \(A^{-1}\)<br>(C) \(A^T A^{-1}\)<br>(D) \(A^2\)</p><p><b>Column II:</b><br>(p) \(\begin{bmatrix} 1-\tan^2 x & 2\tan x \\ -2\tan x & -\tan^2 x+1 \end{bmatrix}\)<br>(q) \(\begin{bmatrix} \dfrac{1}{1+\tan^2 x} & \dfrac{-\tan x}{1+\tan^2 x} \\ \dfrac{\tan x}{1+\tan^2 x} & \dfrac{1}{1+\tan^2 x} \end{bmatrix}\)<br>(r) \(\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}\)<br>(s) \(\begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: For a 2×2 matrix, adj(A) = transpose of cofactor matrix, and det(A) = 1 + tan²x = sec²x. Recognize that A^T A^(-1) produces a rotation matrix when simplified using the identity 1 + tan²x = sec²x and double angle formulas.
<p><strong>Step 1: Calculate det(A)</strong></p><p>det(A) = 1(1) - (tanx)(-tanx) = 1 + tan²x = sec²x</p><p><strong>Step 2: Find adj(A) [Match A→r]</strong></p><p>For 2×2 matrix, adj(A) = transpose of cofactor matrix = [1, -tanx; tanx, 1]</p><p><strong>Step 3: Find A⁻¹ [Match B→q]</strong></p><p>A⁻¹ = adj(A)/det(A) = [1/(1+tan²x), -tanx/(1+tan²x); tanx/(1+tan²x), 1/(1+tan²x)]</p><p><strong>Step 4: Find A^T A⁻¹ [Match C→s]</strong></p><p>A^T = [1, -tanx; tanx, 1]</p><p>A^T A⁻¹ = [1, -tanx; tanx, 1] × [cos²x, -sinxcosx; sinxcosx, cos²x]</p><p>Using cos(2x) = cos²x - sin²x = (1-tan²x)/(1+tan²x) and sin(2x) = 2sinxcosx = 2tanx/(1+tan²x):</p><p>A^T A⁻¹ = [cos(2x), -sin(2x); sin(2x), cos(2x)]</p><p><strong>Step 5: Find A² [Match D→p]</strong></p><p>A² = [1, tanx; -tanx, 1] × [1, tanx; -tanx, 1] = [1-tan²x, 2tanx; -2tanx, 1-tan²x]</p><p>∴ Answer: A→r, B→q, C→s, D→p</p>
Correct Answer: A→r, B→q, C→s, D→p