Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{ccc}1& a& a\\ 0& 1& b\\ 0& 0& 1\end{array}\right],a,b\in \mathbb{R}$. If for some $n\in \mathbb{N}$, $A^n=\left[\begin{array}{ccc}1& 48& 2160\\ 0& 1& 96\\ 0& 0& 1\end{array}\right]$, then $n+a+b$ is equal to

Welcome, fellow traveler on the JEE Advanced journey. Today, we are not just solving a matrix problem; we are uncovering the hidden symmetry within linear transformations.

When you first look at a matrix like $A=\left[\begin{array}{ccc}1& a& a\\ 0& 1& b\\ 0& 0& 1\end{array}\right]$, it might look like a simple grid of numbers. But to the trained eye, this is a structure waiting to reveal its secrets.

We are tasked with finding $n,a,$ and $b$ given that:

When we face powers of matrices, our first instinct should be to look for a pattern. Let us calculate $A^2=A\cdot A$. Performing the row-by-column multiplication, we get:

Notice the beauty here. The diagonal remains $1$. The element in the first row, second column has doubled to $2a$, and the second row, third column has doubled to $2b$.

The top-right element, however, has become $2a+ab$. If we continue to $A^3$, we find:

The coefficients are growing linearly for the off-diagonal elements, while the top-right element is accumulating a more complex sum. This is the heartbeat of the problem.

While induction is a reliable path, there is a more elegant way to view this. Let us decompose $A$ into $I+M$, where $I$ is the identity matrix and $M$ is the strictly upper triangular part:

If you calculate $M^2$, you will find it contains only one non-zero element in the top-right corner, $ab$. Furthermore, $M^3$ is the zero matrix, making $M$ a 'nilpotent' matrix.

Because $I$ and $M$ commute ($IM=MI$), we can use the binomial theorem:

Substituting our matrices back in, we get the general form:

Now, we equate our general form to the given matrix. This gives us three crucial equations:

1) $na=48$

2) $nb=96$

3) $na+\frac{n(n-1)}{2}ab=2160$

Look at the first two equations. If we divide them, the $n$ cancels out, leaving us with $\frac{b}{a}=2$, or $b=2a$.

Now, substitute $b=2a$ and $na=48$ into the third equation:

We can rewrite $n(n-1)a^2$ as $(n-1)(na)a$. Since we know $na=48$, this becomes:

Dividing the entire equation by $48$, we get $1+(n-1)a=45$, which simplifies to $(n-1)a=44$. Expanding this, we have $na-a=44$.

Since $na=48$, we have $48-a=44$, which means $a=4$. With $a=4$, we find $n=\frac{48}{4}=12$, and $b=2(4)=8$.

The final step is simply to sum these values:

We started with a daunting matrix power and ended with a simple sum. This is the essence of JEE Advanced mathematics—taking a complex, intimidating structure and using the right tools to reveal the simple, elegant truth hidden underneath.