Question

If $A = \begin{bmatrix} 2\sin^2 \theta & 0 \\ \sin^2 \theta & \sin^2 \theta \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$ then the number of possible values of $\theta$ in $[-\pi, \pi]$, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

The given matrix is $A = \begin{bmatrix} 2\sin^2 \theta & 0 \\ \sin^2 \theta & \sin^2 \theta \end{bmatrix}$. The inverse of a $2 \times 2$ matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $M^{-1} = \frac{1}{\det(M)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, where $\det(M) = ad - bc$. For the given matrix $A$, $a = 2\sin^2 \theta$, $b = 0$, $c = \sin^2 \theta$, $d = \sin^2 \theta$.

The determinant of A is $\det(A) = (2\sin^2 \theta)(\sin^2 \theta) - (0)(\sin^2 \theta) = 2\sin^4 \theta$. For the inverse $A^{-1}$ to exist, $\det(A) \neq 0$, which means $2\sin^4 \theta \neq 0$, so $\sin \theta \neq 0$.

The inverse of A is $A^{-1} = \frac{1}{2\sin^4 \theta} \begin{bmatrix} \sin^2 \theta & -0 \\ -\sin^2 \theta & 2\sin^2 \theta \end{bmatrix} = \frac{1}{2\sin^4 \theta} \begin{bmatrix} \sin^2 \theta & 0 \\ -\sin^2 \theta & 2\sin^2 \theta \end{bmatrix}$. Since $\sin \theta \neq 0$, $\sin^2 \theta \neq 0$. We can factor out $\sin^2 \theta$ from the matrix: $A^{-1} = \frac{\sin^2 \theta}{2\sin^4 \theta} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \frac{1}{2\sin^2 \theta} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$.

We are given that $A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$. Comparing the calculated inverse with the given inverse, we have: $\frac{1}{2\sin^2 \theta} \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$. This equality holds if and only if the scalar factor is equal to 1: $\frac{1}{2\sin^2 \theta} = 1$. $2\sin^2 \theta = 1$. $\sin^2 \theta = \frac{1}{2}$. Taking the square root, we get $\sin \theta = \pm \frac{1}{\sqrt{2}}$.

We need to find the number of possible values of $\theta$ in the interval $[-\pi, \pi]$ that satisfy $\sin \theta = \frac{1}{\sqrt{2}}$ or $\sin \theta = -\frac{1}{\sqrt{2}}$. Also, we must ensure that $\sin \theta \neq 0$ for these values, which is true for $\pm \frac{1}{\sqrt{2}}$.

Case 1: $\sin \theta = \frac{1}{\sqrt{2}}$. In the interval $[-\pi, \pi]$, the values of $\theta$ for which $\sin \theta = \frac{1}{\sqrt{2}}$ are in the first and second quadrants. The principal value is $\frac{\pi}{4}$. The solutions in $[-\pi, \pi]$ are $\theta = \frac{\pi}{4}$ (first quadrant) and $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ (second quadrant). Both $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are in $[-\pi, \pi]$. There are 2 solutions.

Case 2: $\sin \theta = -\frac{1}{\sqrt{2}}$. In the interval $[-\pi, \pi]$, the values of $\theta$ for which $\sin \theta = -\frac{1}{\sqrt{2}}$ are in the third and fourth quadrants. The reference angle is $\frac{\pi}{4}$. The solutions in $[-\pi, \pi]$ can be found by considering angles in the negative direction from 0. The angle in the fourth quadrant is $-\frac{\pi}{4}$. The angle in the third quadrant is $-\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$. Both $-\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ are in $[-\pi, \pi]$. There are 2 solutions.

The possible values of $\theta$ in $[-\pi, \pi]$ are $\frac{\pi}{4}, \frac{3\pi}{4}, -\frac{\pi}{4}, -\frac{3\pi}{4}$. These are 4 distinct values. For all these values, $\sin \theta = \pm \frac{1}{\sqrt{2}} \neq 0$, so the inverse exists.

The total number of possible values of $\theta$ in $[-\pi, \pi]$ is $2 + 2 = 4$.