Let $A=\begin{bmatrix}1&\sin θ&1\\-\sin θ&1&\sin θ\\-1&-\sin θ&1\end{bmatrix}$, where $0≤θ≤2π$ then which of the following are true?

(A) $|A|= 2+2 \sin^2 θ$
(B) $|A|= 2+ \sin^2 θ$
(C) minimum value of $|A|$ is 1
(D) maximum value of $|A|$ is 4

Choose the correct answer from the options given below:

(A) and (D) only

The correct answer is Option (1) → (A) and (D) only

Given

$A=\begin{pmatrix}1 & \sin\theta & 1\\ -\sin\theta & 1 & \sin\theta\\ -1 & -\sin\theta & 1\end{pmatrix}$

Compute $|A|$.

$|A|=1\begin{vmatrix}1&\sin\theta\\-\sin\theta&1\end{vmatrix} -\sin\theta\begin{vmatrix}-\sin\theta&\sin\theta\\-1&1\end{vmatrix} +1\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix}$

$=1(1+\sin^{2}\theta)-\sin\theta[(-\sin\theta)(1)-(\sin\theta)(-1)] +1[(\sin^{2}\theta)-(-1)]$

$=1+\sin^{2}\theta-\sin\theta(-\sin\theta+\sin\theta)+\sin^{2}\theta+1$

$=1+\sin^{2}\theta+0+\sin^{2}\theta+1$

$|A|=2+2\sin^{2}\theta$

So (A) is true and (B) is false.

Since $0\le \sin^{2}\theta \le 1$:

Minimum $|A|=2$ when $\sin^{2}\theta=0$.

Maximum $|A|=4$ when $\sin^{2}\theta=1$.

Hence (C) is false and (D) is true.

Final answer: (A) and (D)