Animated Solution for Mathematics - Matrices and Determinants: The positive value of the determinant of the matrix $A$, whose $\text{Adj}(\text{Adj}(A))=\left[\begin{array}{ccc}14& 28& -14\\ -14& 14& 28\\ 28& -14& 14\end{array}\right]$, is

Welcome, fellow traveler on the JEE journey. Today, we are going to peel back the layers of a problem that looks like a dense thicket of numbers but is, in reality, a beautiful demonstration of the power of matrix properties.

When you first look at the matrix $M=\text{Adj}(\text{Adj}(A))$, it is natural to feel intimidated. It is a $3\times 3$ grid of integers, and the prospect of calculating the adjoint of an adjoint seems like a nightmare of tedious arithmetic.

But here is the secret: in JEE Advanced, if a problem looks like it requires brute force, you are likely missing an elegant property.

Let us start by grounding ourselves in the theory. We are dealing with the double adjoint of a matrix $A$ of order $n=3$.

There is a powerful identity that acts as our compass here:

Why does this exist? Think of the adjoint as a transformation that maps a matrix to its cofactor space. Applying it twice is a nested transformation.

For a $3\times 3$ matrix, the exponent becomes $(3-1{)}^2=2^2=4$. So, our goal is to find the determinant of the given matrix $M$ and set it equal to $|A{|}^4$. This transforms a daunting matrix problem into a simple algebraic equation: $|A{|}^4=|M|$.

Now, look at the matrix $M$ provided:

Do you see the pattern? Every single element is a multiple of $14$. We can factor out $14$ from the entire matrix.

However, we must be careful. When we factor a scalar $k$ out of a determinant of an $n\times n$ matrix, we are not just pulling it out once; we are pulling it out of every row. Since there are $3$ rows, we pull out $14^3$.

This gives us:

This is the moment where the problem begins to yield. We have reduced the complexity significantly. Now, we only need to calculate the determinant of a simple matrix with small integers.

Let us expand the determinant $\Delta =\left|\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right|$ along the first row. This is a standard procedure, but perform it with focus:

Simplifying this, we get:

So, the determinant of our matrix $M$ is $14^3\times 14$, which is $14^4$. We are almost at the finish line. We established earlier that $|M|=|A{|}^4$.

Therefore, we have the beautiful, symmetric result:

Taking the fourth root, we find $|A|=\pm 14$. The question asks for the positive value, so we conclude that $|A|=14$.

Look at what we just achieved. We didn't calculate a single inverse. We didn't get lost in a sea of cofactors.

By identifying the property $|\text{Adj}(\text{Adj}(A))|=|A{|}^{(n-1{)}^2}$ and correctly applying the scalar property of determinants, we navigated through the complexity with precision.

This is the essence of JEE Advanced mathematics: it is not about how fast you can calculate, but how clearly you can see the underlying structure. Keep this clarity, keep this focus, and you will find that even the most intimidating problems are just puzzles waiting for the right key.