Let $Δ=\begin{bmatrix} 1 & sin θ & 1 \\-sin θ & 1 & sin θ\\-1 & -sin θ & 1\end{bmatrix}$, then Δ lies in the interval

[2, 4]

The correct answer is Option (4) → [2, 4]

$Δ=\begin{bmatrix} 1 & sin θ & 1 \\-sin θ & 1 & sin θ\\-1 & -sin θ & 1\end{bmatrix}$ expanding along with $R_1$

$=1(1+\sin^2θ)+\sin θ(0)+1(\sin^2θ+1)$

$=2+\sin^2θ×2=2(1+\sin^2θ)$

$Δ=2(1+\sin^2θ)$

$0≤\sin^2θ≤1⇒1≤1+\sin^2θ≤2$

$⇒2≤2(1+\sin^2θ)≤4$

$⇒2≤Δ≤4$

$Δ∈[2, 4]$