Let $A = \left| \begin{matrix} 1 & \sin heta & 1 \ - sin heta & 1 & \sin heta \ - 1 & - sin heta & 1 \end{matrix} ight|,\$where $0 \leq heta \leq 2\pi$, then the range of o$|A|$ is

Question

Let $A = \left| \begin{matrix} 1 & \sin	heta & 1 \ - sin	heta & 1 & \sin	heta \ - 1 & - sin	heta & 1 \end{matrix} ight|,\$where $0 \leq 	heta \leq 2\pi$, then the range of o$|A|$ is

Dr. Abubacker · Accepted Answer

(d)
$A = \left| \begin{matrix}
1 & \sin	heta & 1 \
 - sin	heta & 1 & \sin	heta \
 - 1 & - sin	heta & 1
\end{matrix} ight|$
$= 1(1 + \sin^{2}{	heta) -}\sin{	heta( -}\sin{	heta + \sin{	heta) + 1(}}\sin^{2}{	heta + 1)}$
$= 2(1 + \sin^{2}{	heta)}$
Since, the maximum and minimum value of $\sin^{2}	heta$ is 1 and 0.
$	herefore|A| \in \lbrack 2,4brack$

Q. Let $A = \left| \begin{matrix} 1 & \sin\theta & 1 \\ - sin\theta & 1 & \sin\theta \\ - 1 & - sin\theta & 1 \end{matrix} \right|,\$where $0 \leq \theta \leq 2\pi$, then the range of o$|A|$ is

Explanation

Reviewed by Dr. Abubacker

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