<p>We have \[A = \begin{bmatrix}\cos^2\theta & \cos\theta\sin\theta\\ \cos\theta\sin\theta & \sin^2\theta\end{bmatrix}\] and \[B = \begin{bmatrix}\cos^2\phi & \cos\phi\sin\phi\\ \cos\phi\sin\phi & \sin^2\phi\end{bmatrix}.\] If \(AB = O\), then \(\theta - \phi\) is an odd multiple of how many degrees?</p>
Step-by-Step Solution
Key Concept: Matrix A is a rank-1 matrix with structure $\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\end{bmatrix}$. For AB = O with non-zero B, the column space of B must be orthogonal to the row space of A, meaning $\begin{bmatrix}\cos\phi\\\sin\phi\end{bmatrix} \perp \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$.
<p><strong>Step 1:</strong> Recognize the structure of matrix A. We can write:</p><p>$A = \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\end{bmatrix}$</p><p>Similarly, $B = \begin{bmatrix}\cos\phi\\\sin\phi\end{bmatrix}\begin{bmatrix}\cos\phi & \sin\phi\end{bmatrix}$</p><p><strong>Step 2:</strong> Compute AB using the factorization:</p><p>$AB = \begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}(\cos\theta\cos\phi + \sin\theta\sin\phi)\begin{bmatrix}\cos\phi & \sin\phi\end{bmatrix}$</p><p>$= (\cos(\theta-\phi))\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}\begin{bmatrix}\cos\phi & \sin\phi\end{bmatrix}$</p><p><strong>Step 3:</strong> For AB = O, we need either the scalar coefficient or the outer product to vanish. Since B is non-zero rank-1:</p><p>$\cos(\theta-\phi) = 0$</p><p><strong>Step 4:</strong> This gives $\theta - \phi = (2n+1)\frac{\pi}{2}$ for integer n, which is an odd multiple of $\frac{\pi}{2}$ radians.</p><p><strong>Step 5:</strong> Convert to degrees: $\frac{\pi}{2}$ radians = 90°</p><p>∴ Answer: 90
Correct Answer: 90