Question

If $A = \begin{bmatrix} 1 & \cos \theta & 1 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{bmatrix}$; then for all $\theta \in (\frac{3\pi}{4}, \frac{5\pi}{4})$, det(A) lies in the interval:

• A. (1, \frac{5}{2}]
• B. (3, 4]
• C. (0, \frac{3}{2}]
• D. (\frac{3}{2}, 3]

Answer 1

Answer: (B)

(3, 4]

Answer 2

The determinant of the matrix $A$ is calculated as follows:

$\det(A) = 1(1 + \cos^2 \theta) - \cos \theta(-\cos \theta + \cos \theta) + 1(\cos^2 \theta + 1) = 2 + 2\cos^2 \theta$

Given $\theta \in (\frac{3\pi}{4}, \frac{5\pi}{4})$, the range of $\cos \theta$ is $(-1, -\frac{1}{\sqrt{2}}]$. Therefore, the range of $\cos^2 \theta$ is $(\frac{1}{2}, 1]$.

Substituting this into the determinant expression:

$2 + 2(\frac{1}{2}) < 2 + 2\cos^2 \theta \le 2 + 2(1)$

$3 < \det(A) \le 4$

Thus, $\det(A) \in (3, 4]$.