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0 votes
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in Determinants by (26.9k points)
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Mark the correct alternative in the following :

Let A = \(\begin{bmatrix} 1 & sin\,\theta & 1 \\[0.3em] -sin\,\theta & 1 &sin\,\theta \\[0.3em] -1 &-sin\,\theta& 1 \end{bmatrix}\) , where 0 ≤ θ ≤ 2π. Then,

A. Det (A) = 0 

B. Det (A) ∈ (2, ∞) 

C. Det (A) ∈ (2, 4) 

D. Det (A) ∈ [2, 4]

1 Answer

+1 vote
by (27.3k points)
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Best answer

Correct answer : (A)

A = 2 [(sin θ)2+1] 0 ≤ (sin θ)2 ≤ 1 

A ∈ 2 [1,2] 

A ∈ [2,4]

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