<p>The product of matrices \(A = \begin{bmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{bmatrix}\) and \(B = \begin{bmatrix}\cos^2\phi & \cos\phi\sin\phi\\ \cos\phi\sin\phi & \sin^2\phi\end{bmatrix}\) is a null matrix if \(\theta - \phi =\)</p>
<p>(1) \(2n\pi,\ n \in \mathbb{Z}\)</p>
<p>(2) \(n\dfrac{\pi}{2},\ n \in \mathbb{Z}\)</p>
<p>(3) \((2n+1)\dfrac{\pi}{2},\ n \in \mathbb{Z}\)</p>
<p>(4) \(n\pi,\ n \in \mathbb{Z}\)</p>
Step-by-Step Solution
Key Concept: Matrix A has rank 1 (all rows are scalar multiples of [cos θ, sin θ]), and AB = 0 requires the column space of B to be orthogonal to the row space of A, which happens when the direction vectors [cos θ, sin θ] and [cos φ, sin φ] are perpendicular.
<p><strong>Step 1:</strong> Recognize the structure of A and B. Matrix A can be written as A = [cos θ, sin θ]ᵀ[cos θ, sin θ] (outer product of a unit direction vector with itself). Similarly, B = [cos φ, sin φ]ᵀ[cos φ, sin φ].</p><p><strong>Step 2:</strong> For the product AB, compute the (1,1) entry: AB₁₁ = cos²θ·cos²φ + cos θ sin θ·cos φ sin φ = cos θ cos φ(cos θ cos φ + sin θ sin φ) = cos θ cos φ·cos(θ - φ).</p><p><strong>Step 3:</strong> Check the (1,2) entry: AB₁₂ = cos²θ·cos φ sin φ + cos θ sin θ·sin²φ = cos θ sin φ(cos θ cos φ + sin θ sin φ) = cos θ sin φ·cos(θ - φ).</p><p><strong>Step 4:</strong> Notice that AB = [cos θ cos φ, cos θ sin φ]ᵀ·cos(θ - φ). For AB = 0 (null matrix), we need cos(θ - φ) = 0 (since the vector factor is non-zero for generic angles).</p><p><strong>Step 5:</strong> Therefore, θ - φ = π/2 + nπ, or specifically <strong>θ - φ = π/2</strong> (or ±π/2 + nπ for integer n).</p><p>∴ Answer: C</p>
Correct Answer: C