Question

The value of $det A$, where
$A= \begin{pmatrix}1&\cos\,\theta&0\\ -\cos\,\theta&1&\cos\,\theta\\ -1&-\cos\,\theta&1\end{pmatrix}$ 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,
$|A|=\begin{vmatrix}1 & \cos \theta & 0 \\-\cos \theta & 1 & \cos \theta \\
-1 & -\cos \theta & 1\end{vmatrix}$
$=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 \leq \cos \theta \leq 1 $
$\Rightarrow 0 \leq \cos ^{2} \theta \leq 1 $
$ \Rightarrow 1 \leq 1+\cos ^{2} \theta \leq 2$
$\Rightarrow 1 \leq|A| \leq 2 $
$\therefore |A| \in[1,2]$