The parameter on which the value of the deter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the determinant be $\Delta$. Apply the column transformation $C_1 	o C_1 + C_3$. Using the formula $\cos(A-B) + \cos(A+B) = 2\cos A \cos B$ a

Vedclass · Accepted Answer

$p$

Vedclass · Answer

(B) Let the determinant be $\Delta$. Apply the column transformation $C_1 	o C_1 + C_3$. Using the formula $\cos(A-B) + \cos(A+B) = 2\cos A \cos B$ and $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$,we get: $C_1 = \begin{bmatrix} 1+a^2 \ 2\cos px \cos dx \ 2\sin px \cos dx \end{bmatrix}$. Now,apply $C_1 	o C_1 - 2\cos dx \cdot C_2$: $C_1 = \begin{bmatrix} 1+a^2 - 2a\cos dx \ 2\cos px \cos dx - 2\cos px \cos dx \ 2\sin px \cos dx - 2\sin px \cos dx \end{bmatrix} = \begin{bmatrix} 1+a^2 - 2a\cos dx \ 0 \ 0 \end{bmatrix}$. Expanding along the first column: $\Delta = (1+a^2 - 2a\cos dx) \cdot [\cos px \sin(p+d)x - \sin px \cos(p+d)x]$. Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\Delta = (1+a^2 - 2a\cos dx) \cdot \sin((p+d)x - px) = (1+a^2 - 2a\cos dx) \sin dx$. Since the final expression $(1+a^2 - 2a\cos dx) \sin dx$ does not contain $p$,the determinant is independent of $p$.

The parameter on which the value of the determinant $\left| \begin{array}{ccc} 1 & a & a^2 \\ \cos(p-d)x & \cos px & \cos(p+d)x \\ \sin(p-d)x & \sin px & \sin(p+d)x \end{array} \right|$ does not depend is:

Similar Questions

The determinant $\left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} \right|$ is a divisor of:

If $x, y, z$ are not equal and $\neq 0, \neq 1$,the value of $\begin{vmatrix} \log x & \log y & \log z \\ \log 2x & \log 2y & \log 2z \\ \log 3x & \log 3y & \log 3z \end{vmatrix}$ is equal to

The value of $\left| \begin{array}{ccc} 41 & 42 & 43 \\ 44 & 45 & 46 \\ 47 & 48 & 49 \end{array} \right| = $

The value of $\left|\begin{array}{lll}1990 & 1991 & 1992 \\ 1991 & 1992 & 1993 \\ 1992 & 1993 & 1994\end{array}\right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module