Question

The value of $detA$, where
$A=\left(\begin{array}{ccc}1& \cos \theta & 0\\ -\cos \theta & 1& \cos \theta \\ -1& -\cos \theta & 1\end{array}\right)$ lies

• A. in the closed interval $[1,2]$
• B. in the closed interval $[0,1]$
• C. in the open interval $(0,1)$
• D. in the open interval $(1,2)$

Answer 1

Answer: (A)

We have,
$|A|=\left|\begin{array}{ccc}1& \cos \theta & 0\\ -\cos \theta & 1& \cos \theta \\ <br/>-1& -\cos \theta & 1\end{array}\right|$
$=1[1-(-\cos \theta )(\cos \theta )]$
$-\cos \theta [-\cos \theta +\cos \theta ]+0\left({\cos }^2\theta +1\right)$
$=1+{\cos }^2\theta$
Now, we know that
$-1\le \cos \theta \le 1$
$\Rightarrow 0\le {\cos }^2\theta \le 1$
$\Rightarrow 1\le 1+{\cos }^2\theta \le 2$
$\Rightarrow 1\le |A|\le 2$
$\therefore |A|\in [1,2]$