Question

Choose the correct answer. Let , where 0 ≤ θ ≤ 2π, then A. Det (A) = 0 B. Det (A) ∈ (2, ∞) C. Det (A) ∈ (2, 4) D. Det (A)∈ [2, 4]

Answer 1

The given matrix is,

A=[ 1 sinθ 1 −sinθ 1 sinθ −1 −sinθ 1 ]

The determinant of the given matrix is,

| A |=1( 1+ sin 2 θ )−sinθ( −sinθ+sinθ )+1( sin 2 θ+1 ) =1+ sin 2 θ+ sin 2 θ− sin 2 θ+1+ sin 2 θ =2( 1+ sin 2 θ )

As it is given that, 0≤θ≤2π, then multiply this inequality by sinθ,

0≤sinθ≤1 0≤ sin 2 θ≤1 1≤1+ sin 2 θ≤2 2≤2( 1+ sin 2 θ )≤4

Thus, it can be observed that | A |∈[ 2,4 ]

Therefore, option (D) is correct.