Animated Solution for Mathematics - Matrices and Determinants: Let $|A|=6$, where A is a $3\times 3$ matrix. If $|\text{adj}(3\text{ adj}(A^2\cdot \text{adj}(2A)))|=2^m\cdot 3^n,m,n\in \mathbb{N}$, then $m+n$ is equal to .........

A Journey Through Determinants

Welcome, future engineer. Today, we are not just solving a matrix problem; we are peeling an onion. When you look at an expression like $|\text{adj}(3\text{adj}(A^2\cdot \text{adj}(2A)))|$, it is natural to feel a surge of intimidation.

It looks like a chaotic jumble of operators. But here is the secret: mathematics, especially linear algebra, is rarely about brute force. It is about recognizing patterns and respecting the hierarchy of operations.

The Inner Core

We must start at the very heart of the expression: $|2A|$. We are given that $|A|=6$ and the matrix is $3\times 3$.

Recall our first golden rule: when a scalar $k$ is pulled out of a determinant of an $n\times n$ matrix, it emerges as $k^n$. Here, $n=3$ and $k=2$.

We have successfully peeled the first layer! To make our future calculations easier, let us write this in prime factors: $48=2^4\cdot 3^1$.

The Adjoint Transformation

Now, we move outward to $|\text{adj}(2A)|$. We rely on the property that $|\text{adj}(M)|=|M{|}^{n-1}$. Since our matrix is $3\times 3$, the exponent is $3-1=2$.

Substituting our previous result, we get:

The structure is beginning to emerge, and the powers of two and three are starting to align.

The Product Rule

Next, we encounter the term $|A^2\cdot \text{adj}(2A)|$. Let us define this as $|B|$. Using the property $|XY|=|X||Y|$, we split this into $|A^2|\cdot |\text{adj}(2A)|$.

We know $|A^2|=|A{|}^2=6^2=(2\cdot 3{)}^2=2^2\cdot 3^2$. Multiplying this by our result from Phase 2, we get:

This is the core of our remaining expression.

The Final Layers

We are nearing the end. The next layer is $|\text{adj}(B)|$. Applying the adjoint property again, we square our previous result:

Now, we introduce the scalar $3$ inside the adjoint: $|3\text{adj}(B)|$. Again, the scalar $3$ comes out as $3^3$ because the matrix is $3\times 3$.

Finally, we reach the outermost shell: the determinant of the adjoint of this entire expression. We apply the adjoint property one last time:

We are given that this final value equals $2^m\cdot 3^n$. By comparing our result $2^{40}\cdot 3^{22}$ to the given form, we identify $m=40$ and $n=22$.

The final sum is:

Look at what we have achieved. By breaking a complex, nested problem into small, manageable steps, we turned a daunting expression into a simple arithmetic exercise. This is the essence of JEE Advanced preparation: discipline, property knowledge, and the courage to take it one step at a time. Well done.