Find the matrix a of the linear transformation t from ℝ2 to ℝ2 that rotates any vector through an angle of 150∘ in the clockwise direction.

Answer:[tex]A=\left[\begin{array}{cc}-\frac{\sqrt{3} }{2} &\frac{1}{2} \\-\frac{1}{2} &-\frac{\sqrt{3} }{2}\end{array}\right][/tex]Step-by-step explanation:A counterclockwise rotation through an angle [tex]\theta[/tex], can be described with the help of trigonometric functions.The matrix: [tex]\left[\begin{array}{cc}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{array}\right][/tex] describes the counterclockwise rotation of the [tex]R^2[/tex] plane by an angle of [tex]\theta[/tex].For clockwise rotation by an angle of [tex]\theta[/tex] about the origin, we replace [tex]\theta[/tex] by  [tex]-\theta[/tex] to obtain:[tex]\left[\begin{array}{cc}\cos (-\theta)&-\sin (-\theta)\\\sin (-\theta)&\cos (-\theta)\end{array}\right][/tex] Apply the odd and even properties of the sine and cosine functions to obtain:[tex]\left[\begin{array}{cc}\cos (\theta)&\sin (\theta)\\-\sin (\theta)&\cos (\theta)\end{array}\right][/tex].The matrix that describes a rotation of the [tex]R^2[/tex] plane around the origin of [tex]150\degree[/tex] clockwise is:[tex]\left[\begin{array}{cc}\cos (150\degree)&\sin (150\degree)\\-\sin (150\degree)&\cos (150\degree)\end{array}\right]=\left[\begin{array}{cc}-\frac{\sqrt{3} }{2} &\frac{1}{2} \\-\frac{1}{2} &-\frac{\sqrt{3} }{2}\end{array}\right][/tex].Therefore the required matrix is:[tex]A=\left[\begin{array}{cc}-\frac{\sqrt{3} }{2} &\frac{1}{2} \\-\frac{1}{2} &-\frac{\sqrt{3} }{2}\end{array}\right][/tex]The corresponding linear transformation is:[tex]T(x,y)=(-\frac{\sqrt{3} }{2}x +\frac{1}{2}y,\frac{1}{2}x- \frac{\sqrt{3} }{2}y)[/tex]