Question

Let $A=\left[\begin{array}{ccc}1&sin\theta&1\\-sin\theta&1&sin\theta\\-1&-sin\theta&1\end{array}\right]$ , where 0 ≤ θ ≤ 2π. Then
A. Det (A) = 0
B. Det (A) є (2, [infinity])
C. Det (A) є (2, 4)
D. Det (A) є [2, 4]

Answer 1

$A = \left[\begin{array}{ccc}1&sin\theta&1\\-sin\theta&1&sin\theta\\-1&-sin\theta&1\end{array}\right]$

C₁ ----> C₁ + C₂

$A = \left[\begin{array}{ccc}2&sin\theta&1\\0&1&sin\theta\\0&-sin\theta&1\end{array}\right]$

Expanding along C₁ :

= 2 | 1           sinθ |

      | -sinθ      1     |

= 2 [ 1 - (-sin²θ) ]

= 2 [ 1 + sin²θ ]

⇒ 0 ≤ sinθ ≤ 2π

0 ≤ sin²θ ≤ 2π

1 ≤ 1 + sin²θ ≤ 2π   { Adding 1 }

2 ≤ 2(1 + sin²θ) ≤ 4π    { Multiplying 2 }

Hence, Det ( A ) ∈ [2,4]  

(D) is correct option.