Let \(A = \begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\\cos\theta\sin\theta&\sin^2\theta\end{bmatrix}\) and \(B = \begin{bmatrix}\cos^2\phi&\cos\phi\sin\phi\\\cos\phi\sin\phi&\sin^2\phi\end{bmatrix}\). If \(\theta = \displaystyle\frac{\pi}{6}\), \(0 \lt \phi \lt \pi\) and AB is the zero matrix, then \(\phi\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2\pi}{3}\)
The element in row 1 and column 1 of AB is
\(\cos^2\theta\cos^2\phi + \cos\theta\sin\theta\cos\phi\sin\phi\)
\(= \cos\theta\cos\phi\,(\cos\theta\cos\phi + \sin\theta\sin\phi)\)
\(= \cos\theta\cos\phi\cos(\theta - \phi)\)
In the same way every element has the common factor \(\cos(\theta - \phi)\):
\( \displaystyle AB = \cos(\theta - \phi)\begin{bmatrix}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{bmatrix} \)
The matrix in the bracket is never the zero matrix, because the squares of its four elements add to \((\cos^2\theta + \sin^2\theta)(\cos^2\phi + \sin^2\phi) = 1\). So
\(AB = O\) exactly when \(\cos(\theta - \phi) = 0\)
With \(\theta = \displaystyle\frac{\pi}{6}\) and \(0 \lt \phi \lt \pi\), the difference \(\theta - \phi\) lies between \(-\displaystyle\frac{5\pi}{6}\) and \(\displaystyle\frac{\pi}{6}\), so
\( \displaystyle \frac{\pi}{6} - \phi = -\frac{\pi}{2} \)
\( \displaystyle \phi = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{3}\) — Incorrect. \(\cos(\theta - \phi) = \cos\left(-\displaystyle\frac{\pi}{6}\right) \ne 0\).
Option (2) \(\displaystyle\frac{\pi}{2}\) — Incorrect. \(\cos(\theta - \phi) = \cos\left(-\displaystyle\frac{\pi}{3}\right) = \displaystyle\frac{1}{2} \ne 0\).
Option (4) \(\displaystyle\frac{5\pi}{6}\) — Incorrect. \(\cos(\theta - \phi) = \cos\left(-\displaystyle\frac{2\pi}{3}\right) = -\displaystyle\frac{1}{2} \ne 0\).