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) det A=0
b) det $Aϵ(2,\infty )$
c) det $Aϵ(2,4)$
d) det $A\epsilon [2,4]$

Answer 1

Given $A=\left[\begin{array}{ccc}1& sin\theta & 1\\ -sin\theta & 1& sin\theta \\ -1& -sin\theta & 1\end{array}\right]$
$|A|=\left[\begin{array}{ccc}1& sin\theta & 1\\ -sin\theta & 1& sin\theta \\ -1& -sin\theta & 1\end{array}\right]$
$=1(1+si{n}^2\theta )-sin\theta (-sin\theta +sin\theta )+1(si{n}^2\theta +1)=2+2si{n}^2\theta$
For $0\le \theta \le 2\pi ,$
$-1\le sin\theta \le 1\Rightarrow 0\le si{n}^2\theta \le 1$
$\Rightarrow 1\le 1+si{n}^2\theta \le 2\Rightarrow 2\le 2(1+si{n}^2\theta )\le 4$
$\therefore det\left(A\right)\epsilon [2,4]$
Hence, the correct option is (d).