What is the <strong>MINIMUM</strong> value of determinant of A?
Consider the following for the next three (03) items that follow : Let A=\begin{pmatrix}0&\sin^{2}\theta&\cos^{2}\theta\\ \cos^{2}\theta&0&\sin^{2}\theta\\ \sin^{2}\theta&\cos^{2}\theta&0\end{pmatrix} and A=P+Q where P is symmetric matrix and Q is skew-symmetric matrix.
- A. \frac{1}{4} ✓
- B. \frac{1}{2}
- C. \frac{3}{4}
- D. 1
Correct Answer: A. \frac{1}{4}
Explanation
Expanding the determinant |A|, we get \sin^6\theta + \cos^6\theta. This can be rewritten algebraically as 1 - 3\sin^2\theta \cos^2\theta = 1 - \frac{3}{4}\sin^2 2\theta. The minimum value is achieved when \sin^2 2\theta is at its maximum, which is 1. The minimum value is 1 - \frac{3}{4} = \frac{1}{4}.
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