Show that |ˣCᵣ ˣCᵣ₊₁ ˣCᵣ₊₂; ʸCᵣ ʸCᵣ₊₁ ʸCᵣ₊₂; ᶻCᵣ ᶻCᵣ₊₁ ᶻCᵣ₊₂| = |ˣCᵣ ˣ⁺¹Cᵣ₊₁ ˣ⁺²Cᵣ₊₂; ʸCᵣ ʸ⁺¹Cᵣ₊₁ ʸ⁺

Q: Why can't I just expand the determinant directly? It's only a 3x3 matrix.

While you technically could expand it, you would be drowning in a sea of algebraic complexity. The goal of JEE Advanced is not just to solve, but to solve elegantly. Using properties of determinants and binomial identities is the 'architect's approach'—it saves time and prevents calculation errors.

Q: Does the order of column operations matter? Can I do $C_2 \to C_2 + C_1$ first?

Great question! In this specific problem, you have some flexibility, but you must be careful. If you change $C_2$ first, you must ensure that when you subsequently modify $C_3$, you are using the *updated* $C_2$. It is usually safer and more intuitive to work from the rightmost column inwards, as we did in the solution.

Q: How do I know when to stop the operations?

Always keep your eyes on the target! The Right Hand Side (RHS) is your compass. Once your determinant's structure (the upper indices in each column) matches the RHS, you have reached your destination. Don't perform operations for the sake of it; perform them to reach the target state.

Question

Show that $\begin{vmatrix} {}^x C_r & {}^x C_{r+1} & {}^x C_{r+2} \ {}^y C_r & {}^y C_{r+1} & {}^y C_{r+2} \ {}^z C_r & {}^z C_{r+1} & {}^z C_{r+2} \end{vmatrix} = \begin{vmatrix} {}^x C_r & {}^{x+1} C_{r+1} & {}^{x+2} C_{r+2} \ {}^y C_r & {}^{y+1} C_{r+1} & {}^{y+2} C_{r+2} \ {}^z C_r & {}^{z+1} C_{r+1} & {}^{z+2} C_{r+2} \end{vmatrix}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm. Today, we stand before a structure that might look like a wall of intimidating binomial coefficients, but it is actually a beautiful, symmetrical puzzle.

We are looking at a $3 	imes 3$ determinant:

$$\Delta = \begin{vmatrix} {}^{x}C_{r} & {}^{x}C_{r+1} & {}^{x}C_{r+2} \ {}^{y}C_{r} & {}^{y}C_{r+1} & {}^{y}C_{r+2} \ {}^{z}C_{r} & {}^{z}C_{r+1} & {}^{z}C_{r+2} \end{vmatrix}$$

In each row, the upper index is constant—$x$, $y$, or $z$—while the lower index marches forward by one: $r$, $r+1$, $r+2$. This pattern is the heartbeat of the problem.

### The Secret Weapon: Pascal's Identity

Our primary tool is the legendary **Pascal's Identity**:

$${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$$

Think of this as a "merging" operation. It takes two adjacent terms with the same upper index and fuses them into a single term with an incremented upper index.

### The Dance of Operations

Let us begin the transformation. We start by focusing on the third column and applying the operation $C_{3} 	o C_{3} + C_{2}$.

The third column entries become ${}^{x}C_{r+2} + {}^{x}C_{r+1}$. By **Pascal's Identity**, this collapses into ${}^{x+1}C_{r+2}$. We repeat this logic for the $y$ and $z$ rows.

Next, we turn our attention to the second column. We apply $C_{2} 	o C_{2} + C_{1}$.

Now, the second column entries, which were ${}^{x}C_{r+1}$, become ${}^{x}C_{r+1} + {}^{x}C_{r}$. Again, **Pascal's Identity** strikes, simplifying this to ${}^{x+1}C_{r+1}$.

### The Final Push

We are now very close to the target structure. The first column is ${}^{x}C_{r}$, the second is ${}^{x+1}C_{r+1}$, and the third is ${}^{x+1}C_{r+2}$.

To reach the final form, we apply $C_{3} 	o C_{3} + C_{2}$ one last time. The third column becomes ${}^{x+1}C_{r+2} + {}^{x+1}C_{r+1}$.

Applying **Pascal's Identity** for the final time, this becomes ${}^{x+2}C_{r+2}$.

### The Triumph of Elegance

Look at the resulting structure:

$$\Delta = \begin{vmatrix} {}^{x}C_{r} & {}^{x+1}C_{r+1} & {}^{x+2}C_{r+2} \ {}^{y}C_{r} & {}^{y+1}C_{r+1} & {}^{y+2}C_{r+2} \ {}^{z}C_{r} & {}^{z+1}C_{r+1} & {}^{z+2}C_{r+2} \end{vmatrix}$$

We did not brute-force our way through; we danced with the properties of the determinant and the binomial coefficients. This is the essence of **JEE Advanced mathematics**—finding the path of least resistance through the forest of complexity.

Visualized Solution

Understanding the Left Hand Side (LHS)

The Logic Bridge: Pascal's Identity

First Transformation: $C_{3} \to C_{3} + C_{2}$

Simplifying Column 3

Second Transformation: $C_{2} \to C_{2} + C_{1}$

Simplifying Column 2

Final Transformation: $C_{3} \to C_{3} + C_{2}$

The Final Simplification

Conclusion: LHS = RHS

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Secret Weapon

The Dance of Operations

The Final Push

The Triumph of Elegance

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