Solution

1. Key Concepts and Formulas

This problem elegantly combines properties of Arithmetic Progressions (AP) with the manipulation of determinants.

• Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant. If $a, b, c, d$ are in AP with common difference $\lambda$, then:

  • $b = a + \lambda$
  • $c = b + \lambda = a + 2\lambda$
  • $d = c + \lambda = a + 3\lambda$ From these, we can derive useful relationships:
  • $b - a = \lambda$
  • $c - b = \lambda$
  • $d - c = \lambda$
  • $c - a = 2\lambda$
  • $d - b = 2\lambda$
  • $a - c = -2\lambda$
  • $b - d = -2\lambda$

• Properties of Determinants:

  1. Row/Column Operations: The value of a determinant remains unchanged if we apply the operation $R_i \to R_i + kR_j$ or $C_i \to C_i + kC_j$. (Adding a multiple of one row/column to another row/column).
  2. Factoring out a scalar: If all elements of a row or a column are multiplied by a scalar $k$, then the value of the determinant is multiplied by $k$. Conversely, a common factor from any single row or column can be taken out of the determinant.
  3. Expansion: A determinant can be expanded along any row or column. Expanding along a row/column with many zeros simplifies the calculation. For a $3 \times 3$ determinant: \left| {\matrix{ {a_{11}} & {a_{12}} & {a_{13}} \cr {a_{21}} & {a_{22}} & {a_{23}} \cr {a_{31}} & {a_{32}} & {a_{33}} \cr } } \right| = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13} where $C_{ij} = (-1)^{i+j} M_{ij}$ is the cofactor, and $M_{ij}$ is the minor.

2. Problem Setup

We are given that $a, b, c, d$ are in an arithmetic progression with common difference $\lambda$. The value of the determinant is given as 2: D = \left| {\matrix{ {x + a - c} & {x + b} & {x + a} \cr {x - 1} & {x + c} & {x + b} \cr {x - b + d} & {x + d} & {x + c} \cr } } \right| = 2

3. Step-by-Step Solution

Step 1: Simplify the elements of the determinant using AP properties.

• From the AP properties, we know:

  • $a - c = -2\lambda$
  • $d - b = 2\lambda$

• Substitute these into the determinant: D = \left| {\matrix{ {x + (-2\lambda)} & {x + b} & {x + a} \cr {x - 1} & {x + c} & {x + b} \cr {x + (2\lambda)} & {x + d} & {x + c} \cr } } \right| = 2 \Rightarrow D = \left| {\matrix{ {x - 2\lambda} & {x + b} & {x + a} \cr {x - 1} & {x + c} & {x + b} \cr {x + 2\lambda} & {x + d} & {x + c} \cr } } \right| = 2