Question

Choose the correct answer.
Let A=\(\begin{bmatrix}1&sin\theta&1\\-sin\theta&1&sin\theta\\-1&-sin\theta&1\end{bmatrix}\),\(where 0≤\theta≤2\pi,then\)

• A. \(Det(A)=0\)
• B. \(Det (A)∈(2,∞)\)
• C. \(Det(A)∈(2,4)\)
• D. \(Det(A)∈[2,4]\)

Answer 1

The Correct Option is D

A=\(\begin{bmatrix}1&sin\theta&1\\-sin\theta&1&sin\theta\\-1&-sin\theta&1\end{bmatrix}\)
\(∴|A|=1\)\((1+sin^{2}θ)-sinθ(-sinθ+sinθ)+1(sin^{2}θ+1)\)
\(=1+sin^{2}θ+sin^{2}θ+1\)
\(=2+2sin^{2}θ\)
\(=2(1+sin^{2}θ)\)

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

The correct answer is D.