Animated Solution for Mathematics - Matrices and Determinants: If $P=\left[\begin{array}{cc}1& 0\\ 1/2& 1\end{array}\right]$, then $P^{50}$ is:

Mastering Matrix Powers

Imagine you are sitting in the examination hall. The clock is ticking, the pressure is mounting, and you see a question: Find $P^{50}$ where $P=\left[\begin{array}{cc}1& 0\\ \frac{1}{2}& 1\end{array}\right]$.

Your first instinct might be to panic. Do I really have to multiply this matrix fifty times? Will I be here until the exam ends?

Take a deep breath. In the world of JEE Advanced, whenever you see a high power of a matrix, it is rarely a test of your endurance. It is a test of your vision.

Many students fall into the 'brute force' trap. They start multiplying $P\cdot P$, then take that result and multiply by $P$ again.

While this is mathematically sound, it is a strategic error. In competitive exams, we are not just looking for the answer; we are looking for the most elegant path to the truth.

When you see a matrix with simple entries like $1$, $0$, and $\frac{1}{2}$, it is a massive signal that there is a hidden structure waiting to be uncovered. Let us be detectives, not calculators.

Uncovering the Pattern

Let us start by calculating $P^2$. We know that $P^2=P\cdot P$. So, we set up the multiplication:

Performing the row-by-column multiplication, we get:

- First row, first column: $(1)(1)+(0)(\frac{1}{2})=1$ - First row, second column: $(1)(0)+(0)(1)=0$ - Second row, first column: $(\frac{1}{2})(1)+(1)(\frac{1}{2})=\frac{1}{2}+\frac{1}{2}=1$ - Second row, second column: $(\frac{1}{2})(0)+(1)(1)=1$

So, $P^2=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$. To make the pattern obvious, let us write $1$ as $\frac{2}{2}$. Now, $P^2=\left[\begin{array}{cc}1& 0\\ \frac{2}{2}& 1\end{array}\right]$.

Now, let us calculate $P^3$ to confirm our suspicion. $P^3=P^2\cdot P$. Using our result for $P^2$:

Again, the multiplication yields:

- Second row, first column: $(\frac{2}{2})(1)+(1)(\frac{1}{2})=1+\frac{1}{2}=\frac{3}{2}$

So, $P^3=\left[\begin{array}{cc}1& 0\\ \frac{3}{2}& 1\end{array}\right]$. Look at the progression: - For $n=1$, the bottom-left element is $\frac{1}{2}$. - For $n=2$, the bottom-left element is $\frac{2}{2}$. - For $n=3$, the bottom-left element is $\frac{3}{2}$.

Do you see the beauty of it? The matrix is evolving in a perfectly predictable way. The diagonal elements remain constant at $1$, the top-right remains $0$, and the bottom-left element is simply $\frac{n}{2}$.

We can confidently generalize this for any positive integer $n$:

This is the power of mathematical induction. We have moved from a specific calculation to a universal truth for this matrix.

Now, finding $P^{50}$ is trivial. We simply substitute $n=50$ into our general formula:

And there you have it. We didn't need to perform fifty multiplications. We used observation, logic, and the elegance of algebra to dismantle the problem.

This is the mindset of a JEE topper. Never rush into calculation; always pause, observe, and let the pattern reveal itself to you. The final answer is $\left[\begin{array}{cc}1& 0\\ 25& 1\end{array}\right]$.