1. Key Concepts Used

This problem primarily involves the following key mathematical concepts:

• Properties of Determinants: We will extensively use elementary row and column operations to simplify the determinant without changing its value. Specifically, if a determinant $D'$ is obtained from $D$ by applying the operation $R_i \to R_i + kR_j$ (or $C_i \to C_i + kC_j$), then $D' = D$.
• Expansion of a Determinant: To evaluate a determinant, especially after simplification, we expand it along a row or column that contains the maximum number of zeros.
• Trigonometric Identities: While not explicitly used for simplification, the identity $\sin^2\theta + \cos^2\theta = 1$ is implicitly used when we simplify expressions like $\sin^2\theta + \cos^2\theta$.
• General Solution of Trigonometric Equations: For an equation of the form $\cos x = \cos \alpha$, the general solution is given by $x = 2n\pi \pm \alpha$, where $n$ is an integer.
• Domain Restriction: The final solutions must be checked against the given domain for $\theta$.

2. Problem Statement

We are given a determinant equation and asked to find a value of $\theta$ in the interval $\left( {0,{\pi \over 3}} \right)$ that satisfies it. The equation is: \left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0

3. Step-by-Step Solution

Our goal is to simplify the determinant to a simpler algebraic or trigonometric equation, and then solve for $\theta$.

Step 3.1: Simplify the Determinant using Row Operations

The initial determinant looks complex. We can observe common terms in the columns. For instance, the third column has $4\cos 6\theta$ in the first two rows. The first two columns have $\cos^2\theta$ and $\sin^2\theta$ appearing in similar positions. This suggests that row operations will be very effective in creating zeros and simpler terms.

Let the given determinant be $D$. D = \left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0

Apply the following row operations:

1. $R_1 \to R_1 - R_2$: This operation is chosen to simplify the first two elements of the first row and create a zero in the third element.

   • New $R_{11} = (1 + \cos^2\theta) - \cos^2\theta = 1$
   • New $R_{12} = \sin^2\theta - (1 + \sin^2\theta) = -1$
   • New $R_{13} = 4\cos 6\theta - 4\cos 6\theta = 0$

2. $R_2 \to R_2 - R_3$: This operation is chosen to simplify the first two elements of the second row and create a zero in the third element.

   • New $R_{21} = \cos^2\theta - \cos^2\theta = 0$
   • New $R_{22} = (1 + \sin^2\theta) - \sin^2\theta = 1$
   • New $R_{23} = 4\cos 6\theta - (1 + 4\cos 6\theta) = -1$

After these operations, the determinant becomes: \left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0 Notice how we have already introduced two zeros in the first row and one in the second row, which will make expansion much easier.

Step 3.2: Further Simplify using Column Operations

To make the first row even simpler for expansion, we can try to create another zero. Apply the column operation: $C_2 \to C_2 + C_1$: This operation is chosen to create a zero at position $(1,2)$ in the first row.

• New $C_{12} = -1 + 1 = 0$
• New $C_{22} = 1 + 0 = 1$
• New $C_{32} = \sin^2\theta + \cos^2\theta = 1$ (using the identity $\sin^2\theta + \cos^2\theta = 1$)

The determinant now simplifies to: \left| {\matrix{ 1 & 0 & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & 1 & {1 + 4\cos 6\theta } \cr } } \right| = 0 This form is ideal for expansion along the first row ($R_1$) because it contains two zeros.

Step 3.3: Expand the Determinant

Expand the determinant along $R_1$: The expansion formula for a $3 \times 3$ determinant along $R_1$ is $a_{11}(M_{11}) - a_{12}(M_{12}) + a_{13}(M_{13})$, where $M_{ij}$ is the minor corresponding to element $a_{ij}$. 1 \cdot \left| {\matrix{ 1 & { - 1} \cr 1 & {1 + 4\cos 6\theta } \cr } } \right| - 0 \cdot (\text{Minor}) + 0 \cdot (\text{Minor}) = 0 $1 \cdot [1 \cdot (1 + 4\cos 6\theta) - (-1) \cdot 1] = 0$ $1 \cdot (1 + 4\cos 6\theta + 1) = 0$ $2 + 4\cos 6\theta = 0$

Step 3.4: Solve the Trigonometric Equation

Now we have a simple trigonometric equation to solve: $4\cos 6\theta = -2$