Let \(A = \begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}\), where \(0 \le \theta \le 2\pi\) then which of the following are true?
(A) \(|A| = 2 + 2\sin^2\theta\)
(B) \(|A| = 2 + \sin^2\theta\)
(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
Write s for \(\sin\theta\). Expanding the determinant along the first row:
\( |A| = 1\begin{vmatrix}1&s\\-s&1\end{vmatrix} - s\begin{vmatrix}-s&s\\-1&1\end{vmatrix} + 1\begin{vmatrix}-s&1\\-1&-s\end{vmatrix} \)
\(= (1 + s^2) - s(-s + s) + (s^2 + 1)\)
\(= 2 + 2s^2 = 2 + 2\sin^2\theta\)
Since \(0 \le \sin^2\theta \le 1\) on \([0, 2\pi]\):
\(2 \le |A| \le 4\)
The minimum value 2 occurs at \(\theta = 0\) and the maximum value 4 at \(\theta = \frac{\pi}{2}\).
So \(|A| = 2 + 2\sin^2\theta\), its minimum value is 2 and its maximum value is 4.
Checking the statements
(A) \(|A| = 2 + 2\sin^2\theta\) — Correct. This is the expanded determinant.
(B) \(|A| = 2 + \sin^2\theta\) — Incorrect. The coefficient of \(\sin^2\theta\) is 2.
(C) minimum value of \(|A|\) is 1 — Incorrect. The minimum value is 2.
(D) maximum value of \(|A|\) is 4 — Correct. At \(\theta = \frac{\pi}{2}\), \(|A| = 2 + 2 = 4\).
Shortcut: The row operation \(R_3 \to R_3 + R_1\) does not change the determinant and makes the third row \((0, 0, 2)\). Expanding along it:
\( |A| = 2\begin{vmatrix}1&s\\-s&1\end{vmatrix} = 2(1 + s^2) \)