Question

(c) Show that for all values of θ the determinant







1 sin θ 1
− sin θ 1 sin θ
−1 − sin θ 1







lies between 2 and 4 inclusive. State one value of θ for which the determinant has the
value 2, and one for which it has the value 4.

Answer 1

$\begin{vmatrix} 1 & sin \theta & 1\\ -sin \theta & 1 & sin \theta \\ -1& -sin \theta & 1 \end{vmatrix} ​ ​$

$=1\cdot1\cdot1+(-1)\cdot sinθ \cdot sinθ +(- sinθ) \cdot (- sinθ) \cdot 1-$

$-(-1) \cdot 1 \cdot 1 - (-sinθ) \cdot sinθ \cdot 1-$

$- 1 \cdot (-sinθ) \cdot sinθ= 2+2(sinθ)^2$

$-1\le sinθ \le 1 \implies 0\le (sinθ)^2 \le 1$

Then

$2 \le 2+2(sinθ)^2 \le 4$

If $θ=0$ then $2+2(sinθ)^2 =2$

If $θ=π/2$ then $2+2(sinθ)^2 =4$