Animated Solution for Mathematics - Matrices and Determinants: If $A=\left(\begin{array}{cc}0& \sin \alpha \\ \sin \alpha & 0\end{array}\right)$ and $\det (A^2-\frac{1}{2}I)=0$, then a possible value of $\alpha$ is

Imagine you are standing at the threshold of a beautiful geometric transformation. You are given a matrix $A=\left(\begin{array}{cc}0& \sin \alpha \\ \sin \alpha & 0\end{array}\right)$.

At first glance, it looks simple, almost sparse, with those zeros on the main diagonal. In the world of JEE Advanced, simplicity is often a mask for a deeper, more elegant structure.

Our mission is to find the angle $\alpha$ that satisfies the condition $\det (A^2-\frac{1}{2}I)=0$. Let us embark on this journey together.

To understand the behavior of $A^2$, we must perform the matrix multiplication $A\times A$. This is where the magic happens.

When we multiply $A$ by itself, we are essentially applying the transformation twice. Let us set up the multiplication:

As we calculate the top-left element, we take the first row of the first matrix and the first column of the second matrix: $(0\times 0)+(\sin \alpha \times \sin \alpha )={\sin }^2\alpha$.

Now, look at the off-diagonal elements. For the top-right, we have $(0\times \sin \alpha )+(\sin \alpha \times 0)=0$. The same happens for the bottom-left, meaning the off-diagonal elements vanish.

Finally, the bottom-right element is $(\sin \alpha \times \sin \alpha )+(0\times 0)={\sin }^2\alpha$. We are left with a beautiful diagonal matrix:

This is the core geometric reality we needed to visualize.

Now that we have $A^2$, we must subtract $\frac{1}{2}I$. The identity matrix $I$ is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, so $\frac{1}{2}I$ is $\left(\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right)$.

Subtracting this from our diagonal matrix $A^2$ is straightforward:

The determinant of a diagonal matrix is simply the product of its diagonal entries. Thus:

We are given that this determinant is zero. Therefore, ${\left({\sin }^2\alpha -\frac{1}{2}\right)}^2=0$.

If the square of a quantity is zero, the quantity itself must be zero. This leads us to the simple algebraic equation:

Taking the square root of both sides, we find $\sin \alpha =\pm \frac{1}{\sqrt{2}}$. We know from our trigonometric toolkit that $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$.

The final solution for the angle is $\alpha =\frac{\pi }{4}$ (or more generally, $n\pi \pm \frac{\pi }{4}$).

We have navigated the matrix, simplified the determinant, and arrived at the solution with precision. Remember, in physics and math, the most complex-looking problems often collapse into simple, elegant truths if you just take the time to perform the steps with care.