Matrices & Determinants
Matrix Operations
Grade 12
Question:
<p>For each real number x such that -1 < x < 1, let <span>A(x) = ⎡⎣1 -x⎤⎦</span> and let y and z = <span>x/(1-x)</span> and 1/(1-x). Then</p>
<p>(A) A(z) = A(x) + A(y)</p>
<p>(B) A(z) = A(x)[A(y)]⁻¹</p>
<p>(C) A(z) = A(x) × A(y)</p>
<p>(D) A(z) = A(x) - A(y)</p>
Step-by-Step Solution
Key Concept: Recognize that matrix multiplication of A(x) and A(y) yields A(z) where z follows the addition formula for hyperbolic/composition identities
<p><strong>Step 1:</strong> Given <span>A(x) = ⎡⎣1 -x⎤⎦</span>, compute A(y) and A(z)</p><p><strong>Step 2:</strong> First simplify: y = x/(1-x), so 1 + y = 1 + x/(1-x) = 1/(1-x). Also z = x/(1-x) + 1/(1-x) = (x+1)/(1-x)</p><p><strong>Step 3:</strong> Check z directly: z appears to be defined such that we need to verify the relationship.</p><p><strong>Step 4:</strong> Compute A(x)A(y) = ⎡⎣1 -x⎤⎦ · ⎡⎣1 -y⎤⎦ = ⎡⎣1-xy -x-y+xy⎤⎦ = ⎡⎣1-xy -(x+y)/(1-xy)⎤⎦</p><p><strong>Step 5:</strong> Note that if z = (x+y)/(1-xy), then A(z) = ⎡⎣1 -z⎤⎦ = A(x)A(y) when 1-xy ≠ 0</p><p>∴ Answer is C.</p>
Correct Answer: C