Animated Solution for Mathematics - Matrices and Determinants: Which of the following values of $\alpha$ satisfy the equation $\left|\begin{array}{ccc}(1+\alpha {)}^2& (1+2\alpha {)}^2& (1+3\alpha {)}^2\\ (2+\alpha {)}^2& (2+2\alpha {)}^2& (2+3\alpha {)}^2\\ (3+\alpha {)}^2& (3+2\alpha {)}^2& (3+3\alpha {)}^2\end{array}\right|=-648\alpha$?

Taming the Determinant

Welcome, future IITians. Today, we are not just solving a problem; we are embarking on a journey of mathematical refinement.

When you first look at this determinant, it is natural to feel a surge of intimidation. You see quadratic terms, variables, and a three-by-three structure that screams, "Expand me!"

But I want you to pause. Take a deep breath. In the world of JEE Advanced, brute force is rarely the intended path. The problem setters have hidden a beautiful, symmetrical structure within this matrix, and our job is to uncover it.

The Trap of Direct Expansion

Imagine for a moment that you decided to expand this determinant directly. You would be multiplying terms like $(1+\alpha {)}^2(2+2\alpha {)}^2(3+3\alpha {)}^2$.

You would end up with a polynomial of degree six, filled with cross-terms, and the probability of a sign error would be nearly one hundred percent. This is the "Algebraic Nightmare." We must avoid it at all costs.

Instead, we look for patterns. Notice the elements: $(1+\alpha {)}^2,(2+\alpha {)}^2,(3+\alpha {)}^2$. The bases $1,2,3$ are in an arithmetic progression. This is the key. Whenever you see such structure, your intuition should immediately scream: "Row Operations!"

The Surgical Precision of Row Operations

We want to reduce the degree of these quadratic terms. How do we turn a square into a linear term? We use the difference of squares identity: $a^2-b^2=(a-b)(a+b)$.

By performing the row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_2$, we are essentially applying this identity to every single element in the matrix.

Let us look at the first column: $(2+\alpha {)}^2-(1+\alpha {)}^2$. Using our identity, this becomes $(2+\alpha -1-\alpha )(2+\alpha +1+\alpha )$, which simplifies to $(1)(3+2\alpha )$.

Suddenly, the quadratic term has vanished, replaced by a simple linear expression. We repeat this for every column. The determinant, once a monster of squares, is now a sleek, linear matrix:

The Final Simplification

Look at the second and third rows now. They are so similar! The coefficients of $\alpha$ are identical. This is not a coincidence; it is the problem rewarding our insight.

If we perform one more operation, $R_3\to R_3-R_2$, we will eliminate the variable $\alpha$ entirely from the third row.

Subtracting the rows, we get $5-3=2$, $5-3=2$, and $5-3=2$. We are left with a row of twos: $[2,2,2]$. We can factor this $2$ out of the determinant, leaving us with a row of ones.

This is the "Golden State" of determinant problems. With a row of ones, we can easily create zeros using column operations $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$.

The Triumphant Conclusion

Now, the determinant is trivial to expand. We expand along the third row, which is now $[1,0,0]$. The calculation collapses into a simple two-by-two determinant, which simplifies to $-8{\alpha }^3$.

Finally, we equate this to the right-hand side given in the problem:

Rearranging this, we get $8{\alpha }^3-648\alpha =0$, or $8\alpha ({\alpha }^2-81)=0$. The solutions are $\alpha =0,9,-9$.

Look at what we have achieved. We started with a terrifying quadratic matrix and, through the elegance of row and column operations, reduced it to a simple cubic equation. This is the power of mathematical thinking. Never rush to calculate; always pause to simplify. You have mastered the determinant today.