Matrices & Determinants
Transpose and Inverse
Grade 12
Question:
<p>If \(A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}\), then \(A^T A^{-1}\) is</p>
<p>\(\begin{bmatrix} -\cos 2x & \sin 2x \\ -\sin 2x & \cos 2x \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} \cos 2x & \cos 2x \\ \cos 2x & \sin 2x \end{bmatrix}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that A is an orthogonal-like matrix where A^T A = (sec²x)I, allowing us to express A^(-1) = A^T/(sec²x). Then A^T A^(-1) = A^T · A^T/(sec²x) = (A^T)²/(sec²x).
<p><strong>Step 1:</strong> Compute A^T:<br/>A^T = ⎡1 -tan x⎤<br/> ⎣tan x 1⎦</p><p><strong>Step 2:</strong> Compute A^T A:<br/>A^T A = ⎡1 -tan x⎤ ⎡1 tan x⎤ = ⎡1+tan²x 0 ⎤<br/> ⎣tan x 1 ⎦ ⎣-tan x 1 ⎦ ⎣ 0 1+tan²x⎦<br/>= (sec²x)I</p><p><strong>Step 3:</strong> Find A^(-1) from A^T A = (sec²x)I:<br/>A^T A = (sec²x)I → A^(-1) = A^T/(sec²x)</p><p><strong>Step 4:</strong> Compute A^T A^(-1):<br/>A^T A^(-1) = A^T · [A^T/(sec²x)] = (A^T)²/(sec²x)<br/><br/>(A^T)² = ⎡1 -tan x⎤² = ⎡1-tan²x -2tan x⎤<br/> ⎣tan x 1 ⎦ ⎣2tan x 1-tan²x⎦</p><p><strong>Step 5:</strong> Divide by sec²x = (1+tan²x):<br/>A^T A^(-1) = ⎡(1-tan²x)/(1+tan²x) -2tan x/(1+tan²x)⎤<br/> ⎣2tan x/(1+tan²x) (1-tan²x)/(1+tan²x)⎦<br/>= ⎡cos 2x -sin 2x⎤ (using double angle formulas)<br/> ⎣sin 2x cos 2x⎦</p><p>∴ Answer: B</p>
Correct Answer: B