Let A = <table><tr><td>0</td><td>2y</td><td>1</td></tr><tr><td>2x</td><td>y</td><td>-1</td></tr><tr><td>2x</td><td>-y</td><td>1</td></tr></table>. If (x, y ∈ R, x ≠ y) for which A^T A = 3I_3 is :-
Step-by-Step Solution
Key Concept: The condition A^T A = 3I_3 implies that the columns of A are orthogonal and each has a norm of sqrt(3). By calculating the dot products of the columns and setting them to 0, and the squared norms to 3, we can solve for x and y.
Given A = [[0, 2y, 1], [2x, y, -1], [2x, -y, 1]]. A^T A = 3I_3 implies the columns are orthogonal and have length sqrt(3). Column 1: (0, 2x, 2x)^T, Column 2: (2y, y, -y)^T, Column 3: (1, -1, 1)^T. Norm of Col 3: 1^2 + (-1)^2 + 1^2 = 3. Norm of Col 1: 0^2 + (2x)^2 + (2x)^2 = 8x^2 = 3 => x^2 = 3/8. Norm of Col 2: (2y)^2 + y^2 + (-y)^2 = 6y^2 = 3 => y^2 = 1/2. Dot product Col 1 and Col 2: 0 + 2xy - 2xy = 0 (always true). Dot product Col 1 and Col 3: 0 - 2x + 2x = 0 (always true). Dot product Col 2 and Col 3: 2y - y - y = 0 (always true). Since x^2 = 3/8 and y^2 = 1/2, there are 2 possible values for x and 2 for y, giving 4 pairs (x, y).
Correct Answer: 4