<p>Let A = <span><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>0</mn></mtd><mtd><mn>2</mn><mi>y</mi></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn><mi>x</mi></mtd><mtd><mi>y</mi></mtd><mtd><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn><mi>x</mi></mtd><mtd><mo>-</mo><mi>y</mi></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math></span>, (x, y ∈ R, x ≠ y) for which A^T A = 3I_3 is :-</p>
Step-by-Step Solution
Key Concept: Use the condition A^T A = 3I_3 to set up equations for x and y. The matrix product A^T A will result in a diagonal matrix with entries 3, 3, 3. Solving these equations under the constraint x \neq y will yield the number of possible pairs (x, y).
<p>Given A = <span><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>0</mn></mtd><mtd><mn>2</mn><mi>y</mi></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn><mi>x</mi></mtd><mtd><mi>y</mi></mtd><mtd><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn><mi>x</mi></mtd><mtd><mo>-</mo><mi>y</mi></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math></span>. A^T A = 3I_3 implies the columns of A are orthogonal and have length sqrt(3). Calculating A^T A gives: <span><math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>8</mn><msup><mi>x</mi><mn>2</mn></msup></mtd><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd><mtd><mn>6</mn><msup><mi>y</mi><mn>2</mn></msup></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd><mtd><mn>3</mn></mtd></mtr></mtable></mfenced><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>3</mn></mtd><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd><mtd><mn>3</mn></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd><mtd><mn>3</mn></mtd></mtr></mtable></mfenced></math></span>. Thus 8x^2 = 3 => x^2 = 3/8 and 6y^2 = 3 => y^2 = 1/2. This gives x = \pmsqrt(3/8) and y = \pm1/sqrt(2). Since x \neq y, we check the combinations. There are 4 pairs (x, y), but we must exclude cases where x = y. Since sqrt(3/8) \neq 1/sqrt(2), all 4 pairs are valid.</p>
Correct Answer: 2