Question

Let $A=\left[\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right]$, where $0\le \theta \le 2\pi$. Then $|A|$ lies between

Answer 1

$A=\left[\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right]$

$|A|=\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$

$R_1\to R_1+R_3$

$=\left|\begin{array}{ccc}0& 0& 2\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$

$\Rightarrow |A|=2(1+{\sin }^2\theta )$

Since, $0\le \theta \le 2\pi$

$\Rightarrow -1\le \sin \theta \le 1$

$\Rightarrow 0\le {\sin }^2\theta \le 1$

$\Rightarrow 1\le 1+{\sin }^2\theta \le 2$

$\Rightarrow 2\le 2(1+{\sin }^2\theta )\le 4$

$\Rightarrow 2\le |A|\le 4$