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 A, det(A), is calculated as follows:

$\text{det}(A) = 1 \cdot \begin{vmatrix} 1 & \cos \theta \\ -\cos \theta & 1 \end{vmatrix} - \cos \theta \cdot \begin{vmatrix} -\cos \theta & \cos \theta \\ -1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -\cos \theta & 1 \\ -1 & -\cos \theta \end{vmatrix}$

$\text{det}(A) = 1 \cdot (1 + \cos^2 \theta) - \cos \theta \cdot (-\cos \theta + \cos \theta) + 1 \cdot (\cos^2 \theta + 1)$

$\text{det}(A) = (1 + \cos^2 \theta) - \cos \theta \cdot (0) + (1 + \cos^2 \theta)$

$\text{det}(A) = 2 + 2 \cos^2 \theta$

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 $(\frac{1}{2}, 1]$.

Finally, the range of det(A) = $2 + 2 \cos^2 \theta$ is $(3, 4]$.