Matrices & Determinants
Matrix multiplication
Grade 12
Question:
<p>For each real \(x\), \(-1 < x < 1\). Let \(A(x)\) be the matrix \((1-x)^{-1}\begin{bmatrix}1 & -x\\ -x & 1\end{bmatrix}\) and \(z = \dfrac{x+y}{1+xy}\). Then</p>
<p>(1) \(A(z) = A(x)\,A(y)\)</p>
<p>(2) \(A(z) = A(x) - A(y)\)</p>
<p>(3) \(A(z) = A(x) + A(y)\)</p>
<p>(4) \(A(z) = A(x)\,[A(y)]^{-1}\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is an odd function by expanding the determinant, which forces all even-powered terms to vanish, leaving only odd powers. Use the property that ∫_{-a}^{a} f(x)dx = 0 for odd functions to directly conclude the result.
<p><strong>Step 1:</strong> Check if f(x) is an odd function by computing f(-x).</p><p>Replace x with -x in the determinant:</p><p>f(-x) = |(-x)³ (-x)² + 2 (-x) + 1|</p><p> | -2 (-x) + 1 -5 |</p><p> | (-x)⁵ x² + 2 2(-x)|</p><p>= |-x³ x² + 2 -x + 1|</p><p> |-2 -x + 1 -5 |</p><p> |-x⁵ x² + 2 -2x |</p><p><strong>Step 2:</strong> Factor out -1 from Row 1 and Row 3:</p><p>f(-x) = (-1)(-1)|x³ x² + 2 x - 1|</p><p> |-2 -x + 1 -5 |</p><p> |x⁵ x² + 2 2x|</p><p>= |x³ x² + 2 x - 1|</p><p> |-2 -x + 1 -5 |</p><p> |x⁵ x² + 2 2x |</p><p><strong>Step 3:</strong> Swap Row 1 and Row 3 (introduces factor of -1):</p><p>f(-x) = -|x⁵ x² + 2 2x |</p><p> |-2 -x + 1 -5 |</p><p> |x³ x² + 2 x - 1|</p><p><strong>Step 4:</strong> Reorder back to original form—this equals -f(x).</p><p>Therefore f(x) is an <strong>odd function</strong>.</p><p><strong>Step 5:</strong> Use the property: ∫_{-1}^{1} f(x)dx = 0 for any odd function on a symmetric interval.</p><p>∴ Answer: <strong>A</strong></p>
Correct Answer: A