If A = left| {,begin{array}{*{20}{c}}{sin (θ + α )}&{cos (θ + α )}&1{sin (θ + β )}&amp

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3 and 4 .Determinants and Matrices

medium

If $A = \left| {\,\begin{array}{*{20}{c}}{\sin (\theta + \alpha )}&{\cos (\theta + \alpha )}&1\\{\sin (\theta + \beta )}&{\cos (\theta + \beta )}&1\\{\sin (\theta + \gamma )}&{\cos (\theta + \gamma )}&1\end{array}\,} \right|$ ,then

$A = 0$ for all $\theta $

$A$ is an odd Function of $\theta $

$A = 0$ for $\theta = \alpha + \beta + \gamma $

$A$ is independent of $\theta $

Solution

(d) Given $A = \left| {\,\begin{array}{*{20}{c}}{\sin (\theta + \alpha )}&{\cos (\theta + \alpha )}&1\\{\sin (\theta + \beta )}&{\cos (\theta + \beta )}&1\\{\sin (\theta + \gamma )}&{\cos (\theta + \gamma )}&1\end{array}\,} \right|$
Operate ${R_2} \to {R_2} – {R_1},\,{R_3} \to {R_3} – {R_1}$
$\therefore $$A = \{ \cos (\theta + \gamma ) – \cos (\theta + \alpha )\} $
$\{ \sin (\theta + \beta ) – \sin (\theta + \alpha )\} – \{ \cos (\theta + \beta )$

= $\sin (\beta – \gamma ) – \sin (\beta – \alpha ) – \sin (\alpha – \gamma )$,
which is independent of $\theta $.

Standard 12

Mathematics

The system of equations : $2x\, \cos^2\theta + y\, \sin2\theta – 2\sin\theta = 0$ $x\, \sin2\theta + 2y\, \sin^2\theta = – 2\, \cos\theta$ $x\, \sin\theta – y \cos\theta = 0$ , for all values of $\theta$ , can

normal

If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$, $\lambda x+(\lambda-1) y+(\lambda-4) z=7$, $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions, then $\lambda^2+\lambda$ is equal to

medium

(JEE MAIN-2025)

The system of equations $(\sin\theta ) x + 2z = 0$ , $(\cos\theta ) x + (\sin\theta )y = 0$ , $(\cos\theta )y + 2z = a$ has

normal

Let $D _{ k }=\left|\begin{array}{ccc}1 & 2 k & 2 k -1 \\ n & n ^2+ n +2 & n ^2 \\ n & n ^2+ n & n ^2+ n +2\end{array}\right|$. If $\sum \limits_{ k =1}^n$ $D _{ k }=96$, then $n$ is equal to

hard

(JEE MAIN-2023)

$\left| {\,\begin{array}{*{20}{c}}{a – 1}&a&{bc}\\{b – 1}&b&{ca}\\{c – 1}&c&{ab}\end{array}\,} \right| = $

easy