Let $A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{\sin \theta }&1\\{ - \sin \theta }&1&{\sin \theta }\\{ - 1}&{ - \sin \theta }&1\end{array}} \right],$ where $0 \le \theta \le 2\pi .$ Then,
(A) $Det(A) = 0$
(B) $Det(A) \in (2,\infty )$
(C) $Det(A) \in (2,4)$
(D) $Det(A) \in [2,4]$
Option d is correct
: $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{\sin \theta }&1\\{ - \sin \theta }&1&{\sin \theta }\\{ - 1}&{ - \sin \theta }&1\end{array}} \right|$
Expanding along ${R_1},$
we get
$1(1 + {\sin ^2}\theta ) - \sin \theta ( - \sin \theta + \sin \theta ) + 1({\sin ^2}\theta + 1)$
$= 1 + {\sin ^2}\theta + 1 + {\sin ^2}\theta = 2(1 + {\sin ^2}\theta )$
As, ${\sin ^2}\theta \in [0,1]$
$\Rightarrow$ $1 + {\sin ^2}\theta \in [1,2]$
$\Rightarrow$ $2(1 + {\sin ^2}\theta ) \in [2,4]$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Determinants. Curated by Sachin Sharma. Free for all students.