Key Concepts Used:

This problem involves fundamental operations with matrices and the concept of a determinant. Specifically, we will use:

1. Matrix Multiplication: How to multiply two matrices.
2. Identity Matrix ($I$): A special square matrix where all elements on the main diagonal are 1s and all other elements are 0s. For a $2 \times 2$ matrix, $I = \left( {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}} \right)$.
3. Scalar Multiplication of a Matrix: Multiplying every element of a matrix by a scalar value.
4. Matrix Subtraction: Subtracting corresponding elements of two matrices of the same dimensions.
5. Determinant of a $2 \times 2$ Matrix: For a matrix $M = \left( {\begin{matrix} a & b \\ c & d \\ \end{matrix}} \right)$, its determinant is $\det(M) = ad - bc$.
6. Solving Trigonometric Equations: Finding the values of an angle that satisfy a given trigonometric condition.

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Step-by-Step Solution:

Step 1: Calculate $A^2$

First, we need to compute the square of matrix $A$, which is $A \times A$. Given matrix $A = \left( {\begin{matrix} 0 & {\sin \alpha } \\ {\sin \alpha } & 0 \\ \end{matrix}} \right)$.

$A^2 = A \times A = \left( {\begin{matrix} 0 & {\sin \alpha } \\ {\sin \alpha } & 0 \\ \end{matrix}} \right) \left( {\begin{matrix} 0 & {\sin \alpha } \\ {\sin \alpha } & 0 \\ \end{matrix}} \right)$

Explanation: To multiply two matrices, we take the dot product of rows of the first matrix with columns of the second matrix.

• The element in the first row, first column of $A^2$ is $(0 \times 0) + (\sin \alpha \times \sin \alpha) = \sin^2 \alpha$.
• The element in the first row, second column of $A^2$ is $(0 \times \sin \alpha) + (\sin \alpha \times 0) = 0$.
• The element in the second row, first column of $A^2$ is $(\sin \alpha \times 0) + (0 \times \sin \alpha) = 0$.
• The element in the second row, second column of $A^2$ is $(\sin \alpha \times \sin \alpha) + (0 \times 0) = \sin^2 \alpha$.

Therefore, ${A^2} = \left( {\begin{matrix} {{{\sin }^2}\alpha } & 0 \\ 0 & {{{\sin }^2}\alpha } \\ \end{matrix}} \right)$ Tip: Notice that $A^2$ is a scalar matrix, which means it's a scalar multiple of the identity matrix: $A^2 = (\sin^2 \alpha) I$. This observation can sometimes simplify further calculations.

Step 2: Calculate $A^2 - \frac{1}{2}I$

Next, we need to subtract $\frac{1}{2}I$ from $A^2$. The identity matrix $I$ for $2 \times 2$ matrices is $I = \left( {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}} \right)$. So, $\frac{1}{2}I = \frac{1}{2} \left( {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}} \right) = \left( {\begin{matrix} {{1 \over 2}} & 0 \\ 0 & {{1 \over 2}} \\ \end{matrix}} \right)$.

Now, we perform the subtraction: ${A^2} - {1 \over 2}I = \left( {\begin{matrix} {{{\sin }^2}\alpha } & 0 \\ 0 & {{{\sin }^2}\alpha } \\ \end{matrix}} \right) - \left( {\begin{matrix} {{1 \over 2}} & 0 \\ 0 & {{1 \over 2}} \\ \end{matrix}} \right)$

Explanation: To subtract matrices, we subtract the corresponding elements.

• First row, first column: $\sin^2 \alpha - \frac{1}{2}$.
• First row, second column: $0 - 0 = 0$.
• Second row, first column: $0 - 0 = 0$.
• Second row, second column: $\sin^2 \alpha - \frac{1}{2}$.

Thus, ${A^2} - {1 \over 2}I = \left( {\begin{matrix} {{{\sin }^2}\alpha - {1 \over 2}} & 0 \\ 0 & {{{\sin }^2}\alpha - {1 \over 2}} \\ \end{matrix}} \right)$

Step 3: Evaluate the Determinant and Set it to Zero

The problem states that $\det \left( {{A^2} - {1 \over 2}I} \right) = 0$. Let $M = {A^2} - {1 \over 2}I = \left( {\begin{matrix} {{{\sin }^2}\alpha - {1 \over 2}} & 0 \\ 0 & {{{\sin }^2}\alpha - {1 \over 2}} \\ \end{matrix}} \right)$.

Explanation: For a $2 \times 2$ matrix $\left( {\begin{matrix} a & b \\ c & d \\ \end{matrix}} \right)$, its determinant is $ad - bc$. Here, $a = \sin^2 \alpha - \frac{1}{2}$, $b = 0$, $c = 0$, and $d = \sin^2 \alpha - \frac{1}{2}$.

So, the determinant is: $\det \left( {{A^2} - {1 \over 2}I} \right) = \left( {{{\sin }^2}\alpha - {1 \over 2}} \right) \times \left( {{{\sin }^2}\alpha - {1 \over 2}} \right) - (0 \times 0)$ $\det \left( {{A^2} - {1 \over 2}I} \right) = {\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2}$

Given that the determinant is 0: ${\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2} = 0$

Step 4: Solve the Trigonometric Equation

From the equation ${\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2} = 0$, we take the square root of both sides: $\sin^2 \alpha - {1 \over 2} = 0$

Now, isolate $\sin^2 \alpha$: $\sin^2 \alpha = {1 \over 2}$

To find $\sin \alpha$, take the square root of both sides: $\sin \alpha = \pm \sqrt{{1 \over 2}}$ $\sin \alpha = \pm {1 \over {\sqrt 2 }}$

Explanation: This means $\sin \alpha = \frac{1}{\sqrt{2}}$ or $\sin \alpha = -\frac{1}{\sqrt{2}}$. We need to find a value of $\alpha$ that satisfies this condition.

Common Mistake: Forgetting the $\pm$ when taking the square root. Always consider both positive and negative roots.

Step 5: Identify a Possible Value of $\alpha$ from the Options

We are looking for a possible value of $\alpha$. If $\sin \alpha = \frac{1}{\sqrt{2}}$, then a common angle is $\alpha = \frac{\pi}{4}$ (or $45^\circ$). If $\sin \alpha = -\frac{1}{\sqrt{2}}$, then possible angles include $\alpha = \frac{5\pi}{4}$ (or $225^\circ$) or $\alpha = \frac{7\pi}{4}$ (or $315^\circ$).

Let's check the given options: (A) $\alpha = {\pi \over 4}$: Here, $\sin({\pi \over 4}) = {1 \over {\sqrt 2}}$. This satisfies our condition. (B) $\alpha = {\pi \over 6}$: Here, $\sin({\pi \over 6}) = {1 \over 2}$. This does not satisfy our condition. (C) $\alpha = {\pi \over 2}$: Here, $\sin({\pi \over 2}) = 1$. This does not satisfy our condition. (D) $\alpha = {\pi \over 3}$: Here, $\sin({\pi \over 3}) = {{\sqrt 3 } \over 2}$. This does not satisfy our condition.

Since $\alpha = \frac{\pi}{4}$ is one of the solutions and it is present in the options, it is a possible value for $\alpha$.

The final answer is $\boxed{\text{(A)}}$.

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Summary and Key Takeaway:

This problem effectively tests your ability to perform sequential matrix operations (multiplication, scalar multiplication, subtraction) and then apply the concept of a determinant to form an algebraic equation. The final step involves solving a basic trigonometric equation. The key takeaway is to be meticulous with matrix calculations and remember all properties, especially for determinants and trigonometric functions. Recognizing that $A^2$ is a scalar matrix can make the determinant calculation very straightforward.