Animated Solution for Mathematics - Matrices and Determinants: Let $P=\left(\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right)$ and $Q=[q_{ij}]$ be two $3\times 3$ matrices such that $Q-P^5=I_3$. Then $\frac{q_{21}+q_{31}}{q_{32}}$ is equal to:

Imagine you are staring at matrix $P=\left(\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 9& 3& 1\end{array}\right)$ and the problem asks you to compute $P^5$. Your first instinct might be to reach for your pen and start multiplying $P\times P\times P\times P\times P$.

Stop! That is the path to a headache and a potential calculation error. In the world of JEE Advanced, we don't brute-force; we look for the hidden structure.

Notice that $P$ is a lower triangular matrix with $1$s on the main diagonal. We can decompose $P$ into $P=I+X$, where $I$ is the identity matrix and:

By isolating the identity matrix, we have separated the 'static' part of the matrix from the 'active' part.

Now, let's investigate the behavior of $X$. When we calculate $X^2$, we get:

This is interesting, but watch what happens when we calculate $X^3=X^2\cdot X$. Every single element becomes zero! This is the phenomenon of nilpotency.

Because $X^3=0$, it follows that $X^4$, $X^5$, and all higher powers are also zero. We have turned a problem of infinite complexity into a finite, manageable one.

Since $I$ and $X$ commute, we can apply the Binomial Theorem to expand $P^5=(I+X{)}^5$. The expansion is:

Because $X^3$ and all higher powers are zero, the entire series collapses into just three terms:

Now, we compute the pieces:

Adding these to $I$, we get:

The problem defines $Q-P^5=I_3$, which means $Q=P^5+I$. Adding the identity matrix to our result for $P^5$, we get:

We need the ratio $\frac{q_{21}+q_{31}}{q_{32}}$. Substituting our values:

And there you have it! By recognizing the structure and leveraging the power of nilpotency, we solved a seemingly impossible matrix power problem in just a few elegant steps. The final answer is 10.