Animated Solution for Mathematics - Matrices and Determinants: If $P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$ and $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $Q=PA{P}^T$ and $x=P^T{Q}^{2005}P$ then $x$ is equal to

Unlocking the Power of $2005$

Imagine you are standing before a massive, intimidating wall of numbers. The problem asks you to compute $x=P^T{Q}^{2005}P$, where $Q=PA{P}^T$.

Your first instinct might be to panic. Raising a matrix to the power of $2005$ sounds like a nightmare of endless multiplication.

But take a deep breath. In the world of JEE Advanced, whenever you see a massive exponent, it is rarely a test of your stamina; it is a test of your insight. There is a hidden rhythm, a symphony of cancellations waiting to be discovered.

Our journey begins with the matrix

At first glance, it looks like a standard matrix, but let's test its character. We calculate its transpose, $P^T$, and then perform the product $P^{T}P$.

As we multiply the rows of $P^T$ by the columns of $P$, something magical happens. The terms $\frac{3}{4}$ and $\frac{1}{4}$ add up to $1$, and the off-diagonal terms cancel out to $0$.

We have arrived at the identity matrix $I$. This is our first breakthrough: $P$ is an orthogonal matrix. This means $P^T=P^{-1}$, and this property is the key that will unlock the entire problem.

Now, let's turn our attention to the matrix

This is a classic shear matrix. Let's see how it behaves under exponentiation.

We compute

Then

The pattern is undeniable! By the Principle of Mathematical Induction, we can confidently state that for any positive integer $n$,

Thus,

Now, let's look at the expression $Q^n$. We have $Q=PA{P}^T$.

When we square it, we get $Q^2=(PA{P}^T)(PA{P}^T)$. Notice the middle term: $P^{T}P$.

Because $P$ is orthogonal, $P^{T}P=I$. The expression collapses to $PAIA{P}^T$, which is simply $P{A}^2{P}^T$.

This 'telescoping' effect is beautiful. It means that for any power $n$, $Q^n=P{A}^n{P}^T$.

Finally, we substitute this back into our target expression: $x=P^T{Q}^{2005}P$. Substituting $Q^{2005}=P{A}^{2005}{P}^T$, we get

By the associative property, this is $(P^{T}P)A^{2005}(P^{T}P)$. Since $P^{T}P=I$, the entire expression simplifies to $I\cdot A^{2005}\cdot I$, which is just $A^{2005}$.

We started with a terrifying exponent of $2005$, and through the elegance of matrix properties, we reduced it to a simple shear matrix.

The final answer is

This is the beauty of mathematics: when you look past the complexity, you find a structure that is not only solvable but profoundly elegant. Keep this mindset, and no problem will ever be too big for you.