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
Answer & explanation
Correct answer: option 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 $R_1$
$=1(1+\sin^2θ)+\sin θ(0)+1(\sin^2θ+1)$
$=2+\sin^2θ×2=2(1+\sin^2θ)$
$Δ=2(1+\sin^2θ)$
$0≤\sin^2θ≤1⇒1≤1+\sin^2θ≤2$
$⇒2≤2(1+\sin^2θ)≤4$
$⇒2≤Δ≤4$
$Δ∈[2, 4]$