class 12 maths determinants

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) $Det(A) \in [2,4]$

VAVidaara Admin Asked 8d ago 0 views 0 answers
📘 Determinants NCERT,Misc.Q.19,Page.142 SA

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) $Det(A) \in [2,4]$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

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, ${\sin ^2}\theta \in [0,1]$
$\Rightarrow$ $1 + {\sin ^2}\theta \in [1,2]$
$\Rightarrow$ $2(1 + {\sin ^2}\theta ) \in [2,4]$

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