If the product of matrices \(\begin{bmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{bmatrix}\) and \(\begin{bmatrix}\cos^2\beta & \cos\beta\sin\beta\\ \cos\beta\sin\beta & \sin^2\beta\end{bmatrix}\) is a null matrix, then \(\theta\) and \(\beta\) differ by
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → an odd integral multiple of \(\displaystyle\frac{\pi}{2}\)
The (1, 1) element of the product is
\(\cos^2\theta\cos^2\beta+\cos\theta\sin\theta\cos\beta\sin\beta\)
\(= \cos\theta\cos\beta(\cos\theta\cos\beta+\sin\theta\sin\beta)\)
\(= \cos\theta\cos\beta\cos(\theta-\beta)\)
In the same way every element has the common factor \(\cos(\theta-\beta)\):
\( \displaystyle \text{Product} = \cos(\theta-\beta)\begin{bmatrix}\cos\theta\cos\beta & \cos\theta\sin\beta\\ \sin\theta\cos\beta & \sin\theta\sin\beta\end{bmatrix} \)
The four entries of the last matrix cannot all be 0 (cos θ and sin θ are never both 0, nor are cos β and sin β). So the product is the null matrix exactly when
\(\cos(\theta-\beta) = 0\)
\( \displaystyle \theta-\beta = (2n+1)\frac{\pi}{2},\quad n \text{ an integer} \)
So θ and β differ by an odd multiple of \(\frac{\pi}{2}\). Option (3) also allows even multiples of \(\frac{\pi}{2}\), and Option (2) allows only these; there \(\cos(\theta-\beta) = \pm 1\). Option (1) gives \(\cos(\theta-\beta) = -1\).