The multiplicative inverse of A = begin{bmatr | vedclass

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

Question

(B) To find the multiplicative inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|$:$|A| =

Vedclass · Accepted Answer

$\begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$

Vedclass · Answer

(B) To find the multiplicative inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$.First,calculate the determinant $|A|$:$|A| = (\cos 	heta)(\cos 	heta) - (-\sin 	heta)(\sin 	heta) = \cos^2 	heta + \sin^2 	heta = 1$.Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:$	ext{adj}(A) = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.Therefore,$A^{-1} = \frac{1}{1} \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix} = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.

The multiplicative inverse of $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ is

Similar Questions

Find the adjoint of the matrix $A = \left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1\end{array}\right]$.

Which one of the following statements is true?

If $A = \begin{bmatrix} 1 & a & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 13 & 2 & -7 \\ -3 & b & 2 \\ -2 & 0 & 1 \end{bmatrix}$,then the values of $a$ and $b$ are respectively

If $A = \begin{bmatrix} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{bmatrix}$, then $|(\operatorname{Adj} A)^{-1}| = $

If $A$ is a square matrix of order $3 \times 3$,then $|\operatorname{adj} A| =$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module