Matrices & Determinants Question - None

Question

Let the numbers 2, b and c be in an A.P. and $A = \begin{bmatrix} 1 & 1 & 1 \ 2 & b & c \ 4 & b^2 & c^2 \end{bmatrix}$. If $\det(A) \in [2, 16]$, then $c$ lies in the interval:

Accepted Answer

Step 1: Use the A.P. condition. Since 2, b, c are in A.P.:

b = 2 + d and c = 2 + 2d for common difference d

Step 2: Recognize the matrix structure and compute det(A):

This is a Vandermonde-type determinant: det(A) = (b−2)(c−2)(c−b)

Substituting: det(A) = d · 2d · d = 2d³

Step 3: Apply the constraint det(A) ∈ [2, 16]:

For d > 0: 2 ≤ 2d³ ≤ 16 → 1 ≤ d³ ≤ 8 → 1 ≤ d ≤ 2

For d < 0: 2 ≤ −2d³ ≤ 16 → −8 ≤ d³ ≤ −1 → −2 ≤ d ≤ −1

Step 4: Convert to range of c = 2 + 2d:

When 1 ≤ d ≤ 2: c ∈ [4, 6]

When −2 ≤ d ≤ −1: c ∈ [−2, 0]

∴ Answer: D (c ∈ [−2, 0] ∪ [4, 6])

Question:

Step-by-Step Solution

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