Animated Solution for Mathematics - Matrices and Determinants: If $A$ is a $3\times 3$ matrix and $|A|=2$, then $|3\text{ adj}(|3A|A^2)|$ is equal to

Imagine you are standing before a massive, complex matrix problem. It looks like an onion, doesn't it? Layer upon layer of operators, scalars, and matrices.

The secret to solving these JEE Advanced problems is not to panic but to peel them back, one layer at a time. We are given a $3\times 3$ matrix $A$ with $|A|=2$, and we need to evaluate the expression $|3\text{ adj}(|3A|A^2)|$.

The Scalar $k$

The innermost part of our expression is $|3A|$. This is a scalar multiplication inside a determinant.

We know the property $|cA|=c^n|A|$, where $n$ is the order of the matrix. Since $A$ is a $3\times 3$ matrix, $n=3$.

Let's call this value $k=54$. Now, our expression has become much simpler: $|3\text{ adj}(k{A}^2)|$.

Now, we face the outermost scalar, the $3$ sitting inside the determinant. Again, we use the property $|cB|=c^n|B|$, where $B=\text{adj}(k{A}^2)$.

Since $A$ is $3\times 3$, any adjoint of a matrix derived from $A$ is also $3\times 3$. Thus, the $3$ comes out as $3^3=27$. Our expression is now $27|\text{adj}(k{A}^2)|$.

Next, we tackle the adjoint operator. The property for the determinant of an adjoint is $|\text{adj}(B)|=|B{|}^{n-1}$. With $n=3$, this becomes $|B{|}^2$.

We are almost there! We need to pull the scalar $k$ out of the determinant $|k{A}^2|$. Using the property $|cA|=c^n|A|$, we get:

Substituting this back into our expression, we have:

Now, we substitute $k=54$ and $|A|=2$:

To simplify, we use prime factorization: $27=3^3$ and $54=2\times 3^3$. The expression becomes:

Finally, to express this in terms of base $6$, we split $3^{21}$ into $3^{11}\times 3^{10}$:

The complexity melts away, leaving us with a clean, elegant result of $3^{11}\cdot 6^{10}$. You have successfully navigated the layers of this matrix problem!