Let \( A=\begin{bmatrix}1 & \sin\theta & 1\\-\sin\theta & 1 & \sin\theta\\-1 & -\sin\theta & 1\end{bmatrix} \) ,where \( \theta\in[0,2\pi] \) then which of the following statements are correct ?
A. \( |A|=2+2\,\sin^2\theta \)
B. Maximum value of \( |A| \) is 1.
C. Maximum value of \( |A| \) is −1
D. \( |A|\in[2,4] \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A and D only
Expanding along the first row:
\( \displaystyle |A|=1\begin{vmatrix}1 & \sin\theta\\-\sin\theta & 1\end{vmatrix}-\sin\theta\begin{vmatrix}-\sin\theta & \sin\theta\\-1 & 1\end{vmatrix}+1\begin{vmatrix}-\sin\theta & 1\\-1 & -\sin\theta\end{vmatrix} \)
\( \displaystyle =(1+\sin^2\theta)-\sin\theta(-\sin\theta+\sin\theta)+(\sin^2\theta+1) \)
\( \displaystyle =2+2\sin^2\theta \)
For \( \theta\in[0,2\pi] \), \( 0\le\sin^2\theta\le 1 \), so
\( \displaystyle 2\le|A|\le 4 \)
Both ends are reached, at \( \theta=0 \) and \( \theta=\frac{\pi}{2} \).
So \( |A|=2+2\sin^2\theta \), with least value 2 and maximum value 4.
Checking the statements
(A) \( |A|=2+2\,\sin^2\theta \) — Correct. The expansion gives \( 2+2\sin^2\theta \).
(B) Maximum value of \( |A| \) is 1. — Incorrect. The maximum is 4, at \( \sin^2\theta=1 \).
(C) Maximum value of \( |A| \) is −1 — Incorrect. The maximum is 4; |A| is never less than 2.
(D) \( |A|\in[2,4] \) — Correct. |A| runs from 2 at \( \theta=0 \) to 4 at \( \theta=\frac{\pi}{2} \).