Q. 193.9( 34 Votes )

Let <img width="2 <span>A. Det (A) = 0<br></span><br/><span><span>B. Det (A) є (2, ∞)</span></span><br/><br><span><span>C. Det (A) є (2, 4)</span></span><br/><br><span><span>D. Det (A) є [2, 4]</span></span><br/><br>

Answer :

A =

|A| = 1 (1 × 1 – sin θ × (-sin θ)) – sin θ ((-sin θ) × 1 – (-1) × sin θ) + 1 ((-sin θ) × (-sin θ) – (-1) × 1)


|A| = 1 + sin2θ + sin2θ – sin2θ + sin2θ + 1


|A| = 2 + 2sin2θ


|A| = 2(1 + sin2θ)


Now, 0 ≤ θ ≤ 2π


sin 0 ≤ sin θ ≤ sin 2π


0 ≤ sin2θ ≤ 1


1 + 0 ≤ 1 + sin2θ ≤ 1 + 1


2 ≤ 2(1 + sin2θ) ≤ 4


Det (A) є [2, 4]

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