Key Concepts and Formulas

This problem involves concepts from Arithmetic Progressions (AP) and Determinants.

1. Arithmetic Progression (AP): A sequence of numbers such that the difference between consecutive terms is constant. If $a, b, c$ are in AP, then $b-a = c-b = d$, where $d$ is the common difference. This implies $b = a+d$ and $c = a+2d$.
2. Vandermonde Determinant: A determinant of a matrix where each row consists of powers of a variable. A 3x3 Vandermonde determinant of the form $\left| {\begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{matrix}} \right|$ simplifies to $(y-x)(z-x)(z-y)$. This formula provides a quick way to evaluate the determinant in this problem.

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Step-by-Step Solution

1. Expressing terms of the A.P. in terms of the common difference

We are given that the numbers $2, b, c$ are in an Arithmetic Progression (A.P.). Let $d$ be the common difference of this A.P. By the definition of an A.P.: The second term $b$ is equal to the first term plus the common difference: $b = 2 + d$ The third term $c$ is equal to the second term plus the common difference: $c = b + d$ Substituting the expression for $b$ into the equation for $c$: $c = (2 + d) + d$ $c = 2 + 2d$ So, we have $b = 2+d$ and $c = 2+2d$.

2. Evaluating the Determinant of Matrix A

The given matrix A is: $A = \left[ {\begin{matrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & {{b^2}} & {{c^2}} \end{matrix}} \right]$ We will evaluate its determinant, $\det(A)$, using column operations to simplify it. This is a common strategy for determinants with this structure.

• Apply Column Operations: To create zeros in the first row, we perform the following operations:

  • $C_2 \to C_2 - C_1$ (Replace Column 2 with Column 2 minus Column 1)
  • $C_3 \to C_3 - C_1$ (Replace Column 3 with Column 3 minus Column 1) These operations do not change the value of the determinant. $\det(A) = \left| {\begin{matrix} 1 & 1-1 & 1-1 \\ 2 & b-2 & c-2 \\ 4 & {{b^2}}-4 & {{c^2}}-4 \end{matrix}} \right|$ $\det(A) = \left| {\begin{matrix} 1 & 0 & 0 \\ 2 & b-2 & c-2 \\ 4 & (b-2)(b+2) & (c-2)(c+2) \end{matrix}} \right|$ Explanation: We used the algebraic identity $x^2 - y^2 = (x-y)(x+y)$ for the terms in the third row.

• Expand along the first row: Since the first row now has two zeros