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:
Answer & explanation
Correct answer: option 1
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)