If A = begin{bmatrix} 1 & tan(theta/2) -tan( | vedclass

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

Question

(B) Given $A = \begin{bmatrix} 1 & 	an(	heta/2) \ -	an(	heta/2) & 1 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (1)(1) - (	an(\

Vedclass · Accepted Answer

$\cos^2(	heta/2) \cdot A^T$

Vedclass · Answer

(B) Given $A = \begin{bmatrix} 1 & 	an(	heta/2) \ -	an(	heta/2) & 1 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (1)(1) - (	an(	heta/2))(-	an(	heta/2)) = 1 + 	an^2(	heta/2) = \sec^2(	heta/2)$.Since $AB = I$,$B$ is the inverse of $A$,i.e.,$B = A^{-1}$.The inverse of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $\frac{1}{|A|} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$B = \frac{1}{\sec^2(	heta/2)} \begin{bmatrix} 1 & -	an(	heta/2) \ 	an(	heta/2) & 1 \end{bmatrix}$.Since $\frac{1}{\sec^2(	heta/2)} = \cos^2(	heta/2)$ and the transpose $A^T = \begin{bmatrix} 1 & -	an(	heta/2) \ 	an(	heta/2) & 1 \end{bmatrix}$,we get:$B = \cos^2(	heta/2) \cdot A^T$.

If $A = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ and $AB = I$,then $B = $

Similar Questions

If a matrix $A$ is such that $4A^3 + 2A^2 + 7A + I = O$,then $A^{-1}$ equals

If $A(\operatorname{adj} A)=5 I$ where $I$ is the identity matrix of order $3$,then $|\operatorname{adj} A|$ is equal to

If for the matrix $A$,${A^3} = I$,then ${A^{-1}} = $

If matrix $A = \begin{bmatrix} -1 & 2025 & 2026 \\ 0 & 2 & 2027 \\ 0 & 0 & -1 \end{bmatrix}$, then the sum of all elements in $adj(A^{-1})$ is equal to...

If $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$ and $\text{adj } A = \begin{bmatrix} 5 & x & -2 \\ 1 & 1 & 0 \\ -2 & -2 & y \end{bmatrix}$,then the value of $x+y$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module