<p>Given \(A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix}\) and \(\det(A) \in [2, 16]\). If \(2, b, c\) are in A.P., then \(c\) lies in the interval:</p>
Step-by-Step Solution
Key Concept: Recognize that the given matrix is a Vandermonde matrix structure, and use the A.P. condition (b = (2+c)/2) to express the determinant as a function of a single variable, then apply the determinant range constraint.
<p><strong>Step 1: Use the A.P. condition</strong></p><p>Since 2, b, c are in A.P.: <br>b − 2 = c − b<br>⟹ <strong>b = (2+c)/2</strong></p><p><strong>Step 2: Recognize Vandermonde structure</strong></p><p>For the matrix A with rows [1, 1, 1], [2, b, c], [4, b², c²], the determinant is:<br>det(A) = (b−2)(c−2)(c−b)</p><p><strong>Step 3: Express in terms of c only</strong></p><p>Substitute b = (2+c)/2:<br>• b − 2 = (2+c)/2 − 2 = (c−2)/2<br>• c − b = c − (2+c)/2 = (c−2)/2<br><br>det(A) = [(c−2)/2] · (c−2) · [(c−2)/2]<br>= (c−2)³/4</p><p><strong>Step 4: Apply the determinant range</strong></p><p>Given: 2 ≤ |det(A)| ≤ 16<br><br>2 ≤ |(c−2)³/4| ≤ 16<br><br>8 ≤ |(c−2)³| ≤ 64<br><br>Taking cube roots:<br>2 ≤ |c−2| ≤ 4<br><br>This gives: c ∈ [−2, 0] ∪ [4, 6]</p><p><strong>Step 5: Verify feasibility</strong></p><p>For both intervals, b = (2+c)/2 yields distinct real values ensuring matrix non-singularity.</p><p>∴ Answer: <strong>c ∈ [−2, 0] ∪ [4, 6]</strong> (Option C)</p>
Correct Answer: C