Animated Solution for Mathematics - Matrices and Determinants: If $A=\left[\begin{array}{cc}0& -\tan (\theta /2)\\ \tan (\theta /2)& 0\end{array}\right]$ and $(I_2+A)(I_2-A{)}^{-1}=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$, then $13(a^2+b^2)$ is equal to

Welcome, fellow explorer of the mathematical universe! Today, we are going to unravel a problem that might look like a daunting wall of symbols, but beneath the surface, it is a beautiful, choreographed dance between matrix algebra and trigonometry.

When you first look at the matrix $A=\left[\begin{array}{cc}0& -\tan (\theta /2)\\ \tan (\theta /2)& 0\end{array}\right]$, it is natural to feel a bit overwhelmed. But remember, in the world of JEE Advanced, complexity is often just a mask for hidden symmetry. Let us peel back that mask together.

The first step is to simplify our life. Dealing with $\tan (\theta /2)$ inside a matrix is like trying to solve a puzzle with blurry pieces.

Let us make a clean, sharp substitution. We define $t=\tan (\theta /2)$. Suddenly, our matrix $A$ transforms into something much more manageable:

This is the 'Spark' of our solution. By reducing the complexity, we clear the path for the matrix operations that follow. It is a reminder that in mathematics, how you frame the problem is just as important as how you solve it.

Now that we have our simplified matrix $A$, we need to construct the two key players in our expression: $(I_2+A)$ and $(I_2-A)$. The identity matrix $I_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ is our anchor.

Adding $A$ to $I_2$ is straightforward: we simply add the corresponding elements. We get:

Similarly, subtracting $A$ from $I_2$ gives us:

Notice the subtle sign change? That is the heartbeat of the matrix, and it will be crucial for our next step.

To find $(I_2-A{)}^{-1}$, we need the determinant. The determinant of our $2\times 2$ matrix $(I_2-A)$ is $(1)(1)-(t)(-t)=1+t^2$.

With the determinant in hand, the inverse is just a matter of applying the standard formula:

Now, we multiply $(I_2+A)$ by this inverse. We pull the scalar $\frac{1}{1+t^2}$ to the front to keep our workspace clean. The multiplication of $\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]$ is where we must be precise.

Row by row, column by column, we find the resulting elements: $1-t^2$, $-2t$, $2t$, and $1-t^2$.

Here is the moment of truth. We have our resulting matrix:

When we distribute the scalar, we get:

If you have spent time with your half-angle identities, your eyes should light up right now! These are the classic definitions: $\cos \theta =\frac{1-t^2}{1+t^2}$ and $\sin \theta =\frac{2t}{1+t^2}$. By comparing this to the given form $\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$, we identify $a=\cos \theta$ and $b=\sin \theta$.

The problem asks for $13(a^2+b^2)$. Since $a^2+b^2={\cos }^2\theta +{\sin }^2\theta$, and we know the fundamental identity ${\sin }^2\theta +{\cos }^2\theta =1$, the entire expression collapses into:

We have navigated the matrix, invoked the trigonometry, and arrived at the final answer with elegance. Remember, every complex problem is just a series of simple, logical steps waiting for you to connect them.