If A = begin{bmatrix} cos theta & sin theta | vedclass

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

Question

(A) Given,$A = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.The transpose of matrix $A$ is $A^{\prime} = \be

Vedclass · Accepted Answer

$I$

Vedclass · Answer

(A) Given,$A = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.The transpose of matrix $A$ is $A^{\prime} = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$.Now,calculate the product $A \cdot A^{\prime}$:$A \cdot A^{\prime} = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix} \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$Performing matrix multiplication:$A \cdot A^{\prime} = \begin{bmatrix} (\cos 	heta)(\cos 	heta) + (\sin 	heta)(\sin 	heta) & (\cos 	heta)(-\sin 	heta) + (\sin 	heta)(\cos 	heta) \ (-\sin 	heta)(\cos 	heta) + (\cos 	heta)(\sin 	heta) & (-\sin 	heta)(-\sin 	heta) + (\cos 	heta)(\cos 	heta) \end{bmatrix}$Using the trigonometric identity $\cos^{2} 	heta + \sin^{2} 	heta = 1$:$A \cdot A^{\prime} = \begin{bmatrix} \cos^{2} 	heta + \sin^{2} 	heta & -\sin 	heta \cos 	heta + \sin 	heta \cos 	heta \ -\sin 	heta \cos 	heta + \sin 	heta \cos 	heta & \sin^{2} 	heta + \cos^{2} 	heta \end{bmatrix}$$A \cdot A^{\prime} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$Thus,$A \cdot A^{\prime} = I$,which is the identity matrix.

If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then $A \cdot A^{\prime}$ is

Similar Questions

The matrix $A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ is

If the matrix $A$ is both symmetric and skew-symmetric,then

If $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$,verify that $(AB)^{\prime} = B^{\prime} A^{\prime}$.

If $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,verify that $(A+B)^{\prime} = A^{\prime} + B^{\prime}$.

If $A = \begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix}$ and $A + A^{T} = I$, where $I$ is the $2 \times 2$ identity matrix and $A^{T}$ is the transpose of $A$, then the value of $\theta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module