Question

Let \(d \in R\), and \(A = \left[ {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {{\rm{sin}}\theta } \right) - 2}\\ 1&{\left( {{\rm{sin}}\theta } \right) + 2}&d\\ 5&{\left( {2{\rm{sin}}\theta } \right) - d}&{\left( { - {\rm{sin}}\theta } \right) + 2 + 2d} \end{array}} \right]\), θ ∈ [0, 2π]. If the minimum value of det (A) is 8, then a value of d is:
1. -5
2. -7
3. 2(√2 + 1)
4. 2(√2 + 2)

Answer 1

Correct Answer - Option 1 : -5

From question, the given matrix is:

\(\Rightarrow A = \left[ {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {{\rm{sin}}\theta } \right) - 2}\\ 1&{\left( {{\rm{sin}}\theta } \right) + 2}&d\\ 5&{\left( {2{\rm{sin}}\theta } \right) - d}&{\left( { - {\rm{sin}}\theta } \right) + 2 + 2d} \end{array}} \right]\) 

\(\Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {{\rm{sin}}\theta } \right) - 2}\\ 1&{\left( {{\rm{sin}}\theta } \right) + 2}&d\\ 5&{\left( {2{\rm{sin}}\theta } \right) - d}&{\left( { - {\rm{sin}}\theta } \right) + 2 + 2d} \end{array}} \right|\) 

Applying, R₃ → R₃ – 2R₂ + R₁,

\(\Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} { - 2}&{4 + d}&{\left( {{\rm{sin}}\theta } \right) - 2}\\ 1&{\left( {{\rm{sin}}\theta } \right) + 2}&d\\ 1&0&0 \end{array}} \right|\) 

On expanding along R₃,

⇒ |A| = 1[(4 + d)d – (sin θ + 2)(sin θ – 2)]

⇒ |A| = (d² + 4d – sin² θ + 4)

⇒ |A| = (d² + 4d + 4) – sin² θ

∴ |A| = (d + 2)² – sin² θ

We know that, |A| will be minimum if sin² θ is maximum. i.e., if sin² θ = 1.

From question,

⇒ |A|_min = 8

Now,

⇒ (d + 2)² – 1 = 8

⇒ (d + 2)² = 9

⇒ d + 2 = ± 3

∴ d = 1, -5