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Mathematics Question on Determinants

Choose the correct answer.
Let A=[1sinθ1−sinθ1sinθ−1−sinθ1]\begin{bmatrix}1&sin\theta&1\\\\-sin\theta&1&sin\theta\\\\-1&-sin\theta&1\end{bmatrix},where0≤θ≤2π,thenwhere 0≤\theta≤2\pi,then

A

Det(A)=0Det(A)=0

B

Det(A)∈(2,∞)Det (A)∈(2,∞)

C

Det(A)∈(2,4)Det(A)∈(2,4)

D

Det(A)∈[2,4]Det(A)∈[2,4]

Answer

Det(A)∈[2,4]Det(A)∈[2,4]

Explanation

Solution

A=[1sinθ1−sinθ1sinθ−1−sinθ1]\begin{bmatrix}1&sin\theta&1\\\\-sin\theta&1&sin\theta\\\\-1&-sin\theta&1\end{bmatrix}
∴|A|=1$$(1+sin^{2}θ)-sinθ(-sinθ+sinθ)+1(sin^{2}θ+1)
=1+sin2θ+sin2θ+1=1+sin^{2}θ+sin^{2}θ+1
=2+2sin2θ=2+2sin^{2}θ
=2(1+sin2θ)=2(1+sin^{2}θ)

Now,
0≤θ≤2π0≤\theta≤2\pi
⇒0≤sinθ≤1⇒0≤sin\theta≤1
⇒0≤sin2θ≤1⇒0≤sin2\theta≤1
⇒1≤1+sin2θ≤2⇒1≤1+sin2\theta≤2
⇒1≤1+sin2θ≤2⇒1≤1+sin2\theta≤2
∴Det(A)∈[2,4]∴Det(A)∈[2,4]

The correct answer is D.