If A = begin{bmatrix} 1 & tan x -tan x & 1 e | vedclass

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

Question

(B) Given $A = \begin{bmatrix} 1 & 	an x \ -	an x & 1 \end{bmatrix}$.The determinant $|A| = (1)(1) - (	an x)(-	an x) = 1 + 	an^2 x = \sec^2 x$.T

Vedclass · Accepted Answer

$\begin{bmatrix} \cos 2x & -\sin 2x \ \sin 2x & \cos 2x \end{bmatrix}$

Vedclass · Answer

(B) Given $A = \begin{bmatrix} 1 & 	an x \ -	an x & 1 \end{bmatrix}$.The determinant $|A| = (1)(1) - (	an x)(-	an x) = 1 + 	an^2 x = \sec^2 x$.The transpose $A^T = \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix}$.The inverse $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix}$.Now,$A^T \cdot A^{-1} = \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix} \cdot \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix}$$= \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 - 	an^2 x & -	an x - 	an x \ 	an x + 	an x & -	an^2 x + 1 \end{bmatrix}$$= \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 - 	an^2 x & -2 	an x \ 2 	an x & 1 - 	an^2 x \end{bmatrix}$$= \begin{bmatrix} \frac{1 - 	an^2 x}{1 + 	an^2 x} & \frac{-2 	an x}{1 + 	an^2 x} \ \frac{2 	an x}{1 + 	an^2 x} & \frac{1 - 	an^2 x}{1 + 	an^2 x} \end{bmatrix}$Using trigonometric identities $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$ and $\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$,we get:$A^T \cdot A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \ \sin 2x & \cos 2x \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$,then $A^T \cdot A^{-1} = $

Similar Questions

If $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ and $A^{-1} = KA$,then $K$ is

If $\operatorname{det}(AB)=(\operatorname{det} A)(\operatorname{det} B)$ and $A$ is a non-singular matrix of order $3 \times 3$,then $\operatorname{det}(\operatorname{adj} A)=$

The positive value of the determinant of the matrix $A$,whose $\operatorname{Adj}(\operatorname{Adj}(A)) = \begin{bmatrix} 14 & 28 & -14 \\ -14 & 14 & 28 \\ 28 & -14 & 14 \end{bmatrix}$,is

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$,then the value of the determinant of $A^{-1}$ is

If $X$ is a square matrix of order $3 \times 3$ and $\lambda$ is a scalar,then $adj(\lambda X)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module