Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{ccc}2& 1& 0\\ 1& 2& -1\\ 0& -1& 2\end{array}\right]$. If $|adj(adj(adj2A))|=(16{)}^n$, then $n$ is equal to

Navigating Nested Adjoints

Welcome, fellow traveler of the mathematical landscape. Today, we are standing before a problem that looks like a fortress of symbols: $|adj(adj(adj(2A)))|=(16{)}^n$.

At first glance, it is easy to feel overwhelmed. You see nested adjoints, scalar multiplication, and a power of 16. But let me tell you a secret: this problem is not a fortress; it is a puzzle designed to reward those who understand the elegant, underlying architecture of linear algebra.

The Foundation — Determinant of $A$

Before we can scale the heights of nested adjoints, we must understand the ground we stand on. We are given the matrix $A=\left[\begin{array}{ccc}2& 1& 0\\ 1& 2& -1\\ 0& -1& 2\end{array}\right]$.

Our first task is to find its determinant, $|A|$. Expanding along the first row is our most reliable path. We calculate:

There it is. The determinant of our base matrix is 4. This simple number is the key that will unlock the entire problem. Never underestimate the power of a clean, calculated determinant.

The Scalar Trap

Now, we encounter the term $|2A|$. Many students instinctively pull the 2 out and write $2|A|$. But stop!

Remember the geometric soul of the determinant. A $3\times 3$ matrix represents a transformation in 3D space. When you multiply the matrix by 2, you are stretching the space in all three dimensions. Therefore, the volume (the determinant) scales by $2^3$.

Using the property $|kA|=k^m|A|$, where $m=3$ is the order of the matrix, we get:

To make our lives easier later, let us write this as a power of 2: $|2A|=2^5$. Keeping numbers in their exponential form is a pro-tip for JEE; it prevents messy arithmetic and makes comparisons trivial.

The Nested Adjoint Mystery

Now, we face the beast: $|adj(adj(adj(2A)))|$. This looks terrifying, but it is actually a beautiful application of a single, powerful theorem. We know that for any matrix $X$ of order $m$, the determinant of its adjoint is $|adj(X)|=|X{|}^{m-1}$.

What happens when we nest them?

1. $|adj(X)|=|X{|}^{m-1}$

2. $|adj(adj(X))|=|adj(X){|}^{m-1}=(|X{|}^{m-1}{)}^{m-1}=|X{|}^{(m-1{)}^2}$

3. $|adj(adj(adj(X)))|=|adj(adj(X)){|}^{m-1}=(|X{|}^{(m-1{)}^2}{)}^{m-1}=|X{|}^{(m-1{)}^3}$

Do you see the pattern? For $k$ nested adjoints, the exponent is $(m-1{)}^k$. In our case, $m=3$ and $k=3$. Thus, the exponent is $(3-1{)}^3=2^3=8$.

Our expression simplifies to:

The Final Comparison

We are almost there. We have $|2A|=2^5$ and our exponent is 8. Substituting these in:

The problem tells us this value is equal to $16^n$. We need to bridge the gap between base 2 and base 16. We know that $16=2^4$. Therefore, we can rewrite our result:

Comparing $16^{10}$ with $16^n$, the conclusion is immediate: $n=10$.

Look at what we just achieved. We navigated through scalar properties, determinant definitions, and nested adjoint theorems. We didn't brute-force a single matrix multiplication.

We used the elegance of mathematics to dismantle a complex problem piece by piece. This is the essence of JEE Advanced physics and math: it is not about how hard you work, but how well you understand the tools in your kit. Keep this mindset, and no problem will ever be too intimidating again.