Let a, b, c, d be in arithmetic progression with common difference λ. If |x + a - c, x + b, x + a; x

Q: Why can't we just expand the determinant directly?

Expanding a $3 \times 3$ determinant with variables like $x$ directly is a recipe for algebraic disaster. You will end up with a massive polynomial in $x$ that is hard to manage. The beauty of determinant properties is that they allow us to simplify the structure before we ever touch the expansion. Always look for ways to create zeros or constants first.

Q: How do I know which row or column operation to pick?

Look for patterns. In this problem, the second and third columns have terms like $x+b, x+c, x+d$ and $x+a, x+b, x+c$. Subtracting them is a natural choice because the $x$ terms will cancel out, leaving only the differences of the AP terms. Always aim to eliminate the variable $x$ if it appears in every term of a column or row.

Q: Does the value of $x$ matter?

Notice that $x$ completely disappears during the row and column operations. This is a classic JEE Advanced technique. The variable $x$ is often a distractor. If you find that $x$ cancels out, you are on the right track. The problem is designed to test your understanding of AP and determinant properties, not your ability to solve for $x$.

Question

Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If $\begin{vmatrix} x + a - c & x + b & x + a \ x - 1 & x + c & x + b \ x - b + d & x + d & x + c \end{vmatrix} = 2$, then value of $\lambda^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Arithmetic Progression

The problem states that $a, b, c, d$ are in an Arithmetic Progression (AP) with a common difference of $\lambda$. This provides us with the fundamental relationships:
$b - a = \lambda$, $c - b = \lambda$, and $d - c = \lambda$.

Furthermore, we can derive secondary relationships such as $a - c = -2\lambda$ and $b - d = -2\lambda$. These simple linear relationships serve as the foundation for simplifying the determinant.

### Simplifying the Determinant

Direct expansion of the determinant would lead to an unmanageable algebraic expression. Instead, we apply the column operation $C_2 	o C_2 - C_3$.

By subtracting the third column from the second, the $x$ terms are eliminated. The second column transforms into a uniform column of $\lambda$:
$$
\begin{bmatrix}
\lambda \
\lambda \
\lambda
\end{bmatrix}
$$

We can now factor out $\lambda$ from the second column, leaving behind a column of ones. This creates the ideal structure for further row operations.

### Creating Zeros and Expanding

With a column of ones established, we perform the row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$. These operations effectively eliminate the $x$ terms from the second and third rows.

The determinant now contains two zeros in the second column, allowing for a straightforward expansion along that column. The problem reduces to a simple $2 	imes 2$ determinant calculation.

### Final Calculation

As we perform the expansion, the terms involving $x$ cancel out entirely. We are left with the simplified equation:
$$2\lambda^2 = 2$$

Dividing both sides by $2$, we arrive at the result:
$$\lambda^2 = 1$$

The complexity of the original expression has been stripped away to reveal the underlying truth. In JEE Advanced mathematics, success is achieved not through brute force, but by identifying the correct perspective and applying elegant algebraic transformations.

Animated Solution for Mathematics - Matrices and Determinants: Let $a, b, c, d$ be in arithmetic progression with common difference $λ$ . If $x + a - c x - 1 x - b + d x + b x + c x + d x + a x + b x + c = 2$ , then value of $λ^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the AP Terms

Simplifying the First Column

Applying Column Operation $C_{2} \to C_{2} - C_{3}$

Factoring out $λ$

Creating Zeros: $R_{2} \to R_{2} - R_{1}$

Creating More Zeros: $R_{3} \to R_{3} - R_{1}$

Expanding the Determinant

Solving the Quadratic Equation

Final Calculation for $λ^{2}$

The Sigma Insight: Properties of Determinants

Analyzing the Arithmetic Progression

Simplifying the Determinant

Creating Zeros and Expanding

Final Calculation

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