Matrices And Determinants Ques 40

The parameter on which the value of the determinant

$\begin{vmatrix}1 & a & a^{2} \\ \cos (p-d) x & \cos p x & \cos (p+d) x \\ \sin (p-d) x & \sin p x & \sin (p+d) x \end{vmatrix}$

does not depend upon, is

(1997, 2M)

(a) $a$

(b) $p$

(c) $d$

(d) $x$

Show Answer

Answer:

Correct Answer: 40.(b)

Solution:

Formula:

PROPERTIES OF DETERMINANTS:

  1. Let $\Delta=$ $\begin{vmatrix}1 & a & a^{2} \\ \cos (p-d) x & \cos p x & \cos (p+d) x \\ \sin (p-d) x & \sin p x & \sin (p+d) x \end{vmatrix}$

Applying $C_{1} \rightarrow C_{1}+C_{3}$

$ \begin{aligned} & \Rightarrow \Delta=\begin{vmatrix} 1+a^{2} & a & a^{2} \\ \cos (p-d) x+\cos (p+d) x & \cos p x & \cos (p+d) x \\ \sin (p-d) x+\sin (p+d) x & \sin p x & \sin (p+d) x \end{vmatrix} \end{aligned} $

$ \begin{aligned} & \Rightarrow \Delta=\begin{vmatrix} 1+a^{2} & a & a^{2} \\ 2 \cos p x \cos d x & \cos p x & \cos (p+d) x \\ 2 \sin p x \cos d x & \sin p x & \sin (p+d) x \end{vmatrix} \end{aligned} $

Applying $C_{1} \rightarrow C_{1}-2 \cos d x C_{2}$

$ \Rightarrow \Delta=\begin{vmatrix} 1+a^{2}-2 a \cos d x & a & a^{2} \\ 0 & \cos p x & \cos (p+d) x \\ 0 & \sin p x & \sin (p+d) x \end{vmatrix} $

$\Rightarrow \Delta=\left(1+a^{2}-2 a \cos d x\right)[\sin (p+d) x \cos p x$$ -\sin p x \cos (p+d) x] $

$ \Rightarrow \quad \Delta=\left(1+a^{2}-2 a \cos d x\right) \sin d x $

which is independent of $p$.



Table of Contents

sathee Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on

Please select your preferred language
कृपया अपनी पसंदीदा भाषा चुनें