Question

What degree of rotation is represented on this matrix

Answer 1

Answer:

Option B is correct

the degree of rotation is, $-90^{\circ}$

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

$R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$

Given the matrix: $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

Now, equate the given matrix with standard matrix we have;

$\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ =  $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

On comparing we get;

$\cos \theta = 0$       and $-\sin \theta =1$  

As,we know:

• $\cos \theta = \cos(-\theta)$

• $\sin(-\theta) = -\sin \theta$

$\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})$

we get;

$\theta = -90^{\circ}$

and

$\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})$

we get;

$\theta = -90^{\circ}$

Therefore, the degree of rotation is, $-90^{\circ}$