Animated Solution for Mathematics - Matrices and Determinants: If matrix $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ where $a,b,c$ are real positive numbers, $abc=1$ and $A^{T}A=I$, then find the value of $a^3+b^3+c^3$.

Imagine you are standing before a matrix $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$. At first glance, it looks like a simple arrangement of variables.

In the world of JEE Advanced, this is a circulant matrix—a structure of profound symmetry. We are given that $a,b,c$ are positive real numbers, their product $abc=1$, and most importantly, $A^{T}A=I$.

This condition, $A^{T}A=I$, is the heartbeat of the problem. It tells us that $A$ is an orthogonal matrix. Orthogonal matrices preserve the volume of space, and their determinants are constrained by the geometry of the identity matrix.

Let us take the determinant of both sides of $A^{T}A=I$. Using the property $|AB|=|A||B|$ and the fact that $|A^T|=|A|$, we arrive at $|A{|}^2=1$.

This leaves us with two possibilities: $|A|=1$ or $|A|=-1$. We are at a crossroads and must determine which one is correct by expanding the determinant of $A$ directly.

Expanding along the first row, we get:

Distributing these terms, we find:

Grouping them, we arrive at the beautiful expression:

Now, look at the expression $3abc-(a^3+b^3+c^3)$. It is the negative of the famous algebraic identity $a^3+b^3+c^3-3abc$.

We know that:

By multiplying and dividing by $2$, we can rewrite the second factor as:

Since $a,b,c>0$, the sum $(a+b+c)$ is positive, and the sum of squares is non-negative. Thus, $a^3+b^3+c^3-3abc\ge 0$.

Because $a^3+b^3+c^3-3abc\ge 0$, it follows that $-(a^3+b^3+c^3-3abc)\le 0$. Our determinant $|A|$ is exactly this negative expression.

Therefore, $|A|\le 0$. Combining this with our earlier finding that $|A|=\pm 1$, the only logical conclusion is that $|A|=-1$.

Now, we substitute this back into our equation:

Given $abc=1$, we have $3(1)-(a^3+b^3+c^3)=-1$. Rearranging this, we find:

We have arrived at the destination. The final value is 4.