Let a 0, d 0. Find the value of the determinant |1/a 1/(a(a+d)) 1/((a+d)(a+2d)); 1/(a+d) 1/((a+d

Q: Wait, why do we multiply rows by different factors? Doesn't that change the determinant?

That is a brilliant observation! Yes, multiplying a row by a factor $k$ multiplies the determinant by $k$. To keep the value unchanged, we must divide the determinant by that same factor outside. Think of it as a balancing act: what we do inside, we must undo outside. We are essentially factoring out the common denominators to make the matrix 'clean'.

Q: Why can't we just expand the determinant directly? It looks like a standard 3x3 matrix.

You absolutely could, but you would be drowning in a sea of algebraic terms like $a(a+d)(a+2d)...$ and it would be a nightmare to simplify. In JEE Advanced, we don't just solve; we solve elegantly. Substitution is our tool to reveal the hidden symmetry. If you expand directly, you risk a calculation error that could cost you precious marks.

Q: I got stuck after the first row operation. How do I know which operation to pick?

Great question. Always look for the common difference $d$. When you see terms like $x_k = a+kd$, subtracting them will almost always yield $d$ or a multiple of $d$. When you see $x_3 - x_2 = d$, you know you are on the right path. The goal is to create zeros, and subtracting rows is the most powerful way to do that.

Question

Let $a > 0, d > 0$. Find the value of the determinant $\begin{vmatrix} \frac{1}{a} & \frac{1}{a(a+d)} & \frac{1}{(a+d)(a+2d)} \ \frac{1}{(a+d)} & \frac{1}{(a+d)(a+2d)} & \frac{1}{(a+2d)(a+3d)} \ \frac{1}{(a+2d)} & \frac{1}{(a+2d)(a+3d)} & \frac{1}{(a+3d)(a+4d)} \end{vmatrix}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The first thing that should catch your eye is the structure of the denominators. We have $a$, $(a+d)$, $(a+2d)$, and so on, which forms an Arithmetic Progression. If we try to calculate this determinant as it stands, we will be writing pages of algebra.

Instead, let us introduce a new variable, $x_k = a + kd$. By doing this, we are not just simplifying the math; we are changing our perspective. We are turning a complex algebraic expression into a clean, manageable matrix.

Our determinant $\Delta$ now looks like this:

$$
\Delta = \begin{vmatrix} 
\frac{1}{x_0} & \frac{1}{x_0 x_1} & \frac{1}{x_1 x_2} \ 
\frac{1}{x_1} & \frac{1}{x_1 x_2} & \frac{1}{x_2 x_3} \ 
\frac{1}{x_2} & \frac{1}{x_2 x_3} & \frac{1}{x_3 x_4} 
\end{vmatrix}
$$

Suddenly, the structure is visible. We have a pattern. We have a rhythm.

### The Balancing Act

Fractions in a determinant are like obstacles on a racetrack; they slow you down and increase the chance of a crash. We need to clear them.

To clear the first row, we multiply by $x_0 x_1 x_2$. But wait! We cannot just change the value of the determinant. We must compensate. If we multiply the inside by $x_0 x_1 x_2$, we must divide the outside by $x_0 x_1 x_2$.

We apply this logic to every row. For the second row, we multiply by $x_1 x_2 x_3$, and for the third, by $x_2 x_3 x_4$. The outside factor becomes a massive product:

$$
	ext{Factor} = \frac{1}{x_0 x_1^2 x_2^3 x_3^2 x_4}
$$

It looks intimidating, but keep it safe in your pocket. We will deal with it at the very end. Inside the matrix, the fractions have vanished, leaving us with a beautiful, clean structure:

$$
\Delta' = \begin{vmatrix} 
x_1 x_2 & x_2 & x_0 \ 
x_2 x_3 & x_3 & x_1 \ 
x_3 x_4 & x_4 & x_2 
\end{vmatrix}
$$

### The Symphony of Zeros

Now, we are in the endgame. We have a clean matrix, and our goal is to create zeros. Zeros are the best friends of a determinant because they make expansion trivial.

Let us perform our first operation: $R_2 	o R_2 - R_1$. Look at the elements: $x_2 x_3 - x_1 x_2$ becomes $x_2(x_3 - x_1)$, which is $2d x_2$. The other elements become $x_3 - x_2 = d$ and $x_1 - x_0 = d$.

Our second row is now $[2d x_2, d, d]$. Now, let us perform $R_3 	o R_3 - R_2$. Using the same logic, the third row becomes $[2d x_3, d, d]$.

Do you see the symmetry? If we perform $R_3 	o R_3 - R_2$ one more time, the last two elements of the third row become zero! The first element becomes $2d(x_3 - x_2) = 2d^2$. Our matrix is now:

$$
\begin{vmatrix} 
x_1 x_2 & x_2 & x_0 \ 
2d x_2 & d & d \ 
2d^2 & 0 & 0 
\end{vmatrix}
$$

### The Final Victory

Expanding along the third row is now a piece of cake. We have $2d^2$ multiplied by the minor:

$$
\begin{vmatrix} 
x_2 & x_0 \ 
d & d 
\end{vmatrix}
$$

This simplifies to $2d^2(d x_2 - d x_0) = 2d^3(x_2 - x_0)$. Since $x_2 - x_0 = 2d$, our $\Delta'$ is $4d^4$.

Finally, we bring back our outside factor. The result is:

$$
\frac{4d^4}{x_0 x_1^2 x_2^3 x_3^2 x_4}
$$

Substituting back our original values, we get the final answer:

$$
\mathbf{\frac{4d^4}{a(a+d)^2(a+2d)^3(a+3d)^2(a+4d)}}
$$

You did it. You took a terrifying problem and broke it down into simple, logical steps. This is the essence of JEE Advanced mathematics: it is not about memorizing formulas; it is about seeing the structure, trusting the process, and having the patience to let the solution unfold.

Visualized Solution

The Determinant Structure

Rewriting the Determinant

Clearing Denominators - Row 1

Clearing Denominators - Row 2 & 3

The Simplified Matrix

First Row Operation

Second Row Operation

Creating Zeros

Evaluating the Determinant

Final Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Balancing Act

The Symphony of Zeros

The Final Victory

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