Let $A=\begin{bmatrix}1 & \sin heta & 1\ -\sin heta & 1 & \sin heta\ -1 & -\sin heta & 1\end{bmatrix}$, where $0\leq heta\leq2\pi$. Then:

Question

Let $A=\begin{bmatrix}1 & \sin	heta & 1\ -\sin	heta & 1 & \sin	heta\ -1 & -\sin	heta & 1\end{bmatrix}$, where $0\leq	heta\leq2\pi$. Then:

Aicharya · Accepted Answer

$A=\begin{bmatrix} 1 & \sin\theta & 1 \ -\sin\theta & 1 & \sin\theta \ -1 & -\sin\theta & 1 \end{bmatrix}$ $\therefore|\text{A}|=1(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+1(\sin^2\theta+1)$ $=1+\sin^2\theta+\sin^2\theta+1$ = 2 + 2 sin 2 $\theta$ = 2 (1 + sin 2 $\theta$ ) Now, $0\leq\theta\leq2\pi$ $\Rightarrow 0\leq\sin\theta\leq1$ $\Rightarrow 0\leq1+\sin^2\theta\leq2$ $\Rightarrow2\leq2(1+\sin^2\theta)\leq4$ $\therefore$ Det (A) $\in$ [2, 4] The correct answer is d.

Let $A=\begin{bmatrix}1 & \sin\theta & 1\\ -\sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1\end{bmatrix}$…

Solution