Finding the Minimum Determinant of Matrix A

The problem asks for the minimum value of the determinant of the given matrix A, which involves trigonometric functions.

The given matrix A is:

\(A = \left(\begin{array}{ccc}0 & \sin^2 \theta & \cos^2 \theta \\ \cos^2 \theta & 0 & \sin^2 \theta \\ \sin^2 \theta & \cos^2 \theta & 0\end{array}\right)\)

The information about A = P + Q where P is symmetric and Q is skew-symmetric is a standard matrix decomposition, but it does not directly affect the calculation or value of the determinant of A itself. We will focus on calculating the determinant of A.

Calculating the Determinant of A

We can calculate the determinant of A using the cofactor expansion along the first row:

\(\det(A) = 0 \cdot \det\left(\begin{array}{cc}0 & \sin^2 \theta \\ \cos^2 \theta & 0\end{array}\right) - \sin^2 \theta \cdot \det\left(\begin{array}{cc}\cos^2 \theta & \sin^2 \theta \\ \sin^2 \theta & 0\end{array}\right) + \cos^2 \theta \cdot \det\left(\begin{array}{cc}\cos^2 \theta & 0 \\ \sin^2 \theta & \cos^2 \theta\end{array}\right)\)

Now, we calculate the 2x2 determinants:

• The first term is \(0 \cdot (0 \cdot 0 - \sin^2 \theta \cdot \cos^2 \theta) = 0\).
• The second term is \(-\sin^2 \theta \cdot (\cos^2 \theta \cdot 0 - \sin^2 \theta \cdot \sin^2 \theta) = -\sin^2 \theta \cdot (-\sin^4 \theta) = \sin^6 \theta\).
• The third term is \(\cos^2 \theta \cdot (\cos^2 \theta \cdot \cos^2 \theta - 0 \cdot \sin^2 \theta) = \cos^2 \theta \cdot (\cos^4 \theta) = \cos^6 \theta\).

So, the determinant of A is:

\(\det(A) = 0 + \sin^6 \theta + \cos^6 \theta\)

\(\det(A) = \sin^6 \theta + \cos^6 \theta\)

Simplifying the Determinant using Trigonometric Identities

We can simplify the expression \(\sin^6 \theta + \cos^6 \theta\) using trigonometric identities. We use the sum of cubes formula \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) with \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\):

\(\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3\)
\(= (\sin^2 \theta + \cos^2 \theta) ((\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2)\)

Since \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to:

\(\det(A) = 1 \cdot (\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta)\)
\(= \sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta\)

Next, we simplify \(\sin^4 \theta + \cos^4 \theta\). We know that \(a^2 + b^2 = (a+b)^2 - 2ab\). Let \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\):

\(\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta\)
\(= (1)^2 - 2 \sin^2 \theta \cos^2 \theta\)
\(= 1 - 2 \sin^2 \theta \cos^2 \theta\)

Substitute this back into the expression for \(\det(A)\):

\(\det(A) = (1 - 2 \sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta\)
\(\det(A) = 1 - 3 \sin^2 \theta \cos^2 \theta\)

Using the identity \(\sin(2\theta) = 2 \sin \theta \cos \theta\), we have \(\sin^2(2\theta) = 4 \sin^2 \theta \cos^2 \theta\). Thus, \(\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2\theta)\).

Substitute this into the expression for \(\det(A)\):

\(\det(A) = 1 - 3 \left(\frac{1}{4} \sin^2(2\theta)\right)\)
\(\det(A) = 1 - \frac{3}{4} \sin^2(2\theta)\)

Finding the Minimum Value of the Determinant

We need to find the minimum value of \(\det(A) = 1 - \frac{3}{4} \sin^2(2\theta)\).
We know that the range of \(\sin(2\theta)\) is \([-1, 1]\).

Therefore, the range of \(\sin^2(2\theta)\) is \([0, 1]\).
This means \(0 \le \sin^2(2\theta) \le 1\).

To minimize the expression \(1 - \frac{3}{4} \sin^2(2\theta)\), we need to subtract the largest possible value from 1. This occurs when \(\sin^2(2\theta)\) is at its maximum value, which is 1.

The maximum value of \(\sin^2(2\theta)\) is 1, which happens when \(\sin(2\theta) = 1\) or \(\sin(2\theta) = -1\).

Substituting the maximum value of \(\sin^2(2\theta) = 1\) into the determinant expression:

Minimum \(\det(A) = 1 - \frac{3}{4} \cdot (1)\)
Minimum \(\det(A) = 1 - \frac{3}{4}\)
Minimum \(\det(A) = \frac{4-3}{4}\)
Minimum \(\det(A) = \frac{1}{4}\)

Thus, the minimum value of the determinant of A is \(\frac{1}{4}\).

Revision Table: Determinant Calculation and Simplification

Additional Information: Trigonometric Identities and Ranges

Understanding trigonometric identities and the range of trigonometric functions is crucial for solving problems like this. Key identities used here include:

• Pythagorean Identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Sum of Cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)
• Double Angle Identity: \(\sin(2\theta) = 2 \sin \theta \cos \theta\)

Also important is the range of the square of the sine function. For any real \(\phi\), the range of \(\sin(\phi)\) is \([-1, 1]\). Squaring this value gives a range of \([0, 1]\) for \(\sin^2(\phi)\). In this problem, \(\phi = 2\theta\), so \(0 \le \sin^2(2\theta) \le 1\). Knowing this range allows us to find the maximum and minimum values of expressions involving \(\sin^2(2\theta)\).