If A=[1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1]A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}…

Mathematics

Problem

If A=[1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1]A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix} ; then for all θ∈(3π4,5π4)\theta \in \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) , det⁡(A)\det(A) lies in the interval : (1) [52,4)\left[\frac{5}{2}, 4\right) (2) [32,3]\left[\frac{3}{2}, 3\right] (3) (0,32]\left(0, \frac{3}{2}\right] (4) (1,52]\left(1, \frac{5}{2}\right]

Browse the problem library