91Ó°ÊÓ

A Square matrix is orthogonal, if the product of the square matrix (C)and the transpose of the matrix$\left(C^T\right)$, are equal to the identity matrix.

The transpose of the matrix$\left(C^T\right)$ is the inverse of the square matric (C).

If the determinant of the square matrix is equal to the 1, then the matrix represents a rotation.

If the determinant of the square matrix is equal to the -1, then the matrix represents a reflection.

A matrix$\frac{1}{2}\left[\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right]$is given.

The determinant, rotation angle, or the line of reflection of the given matrix need to be determined and also prove that it is orthogonal.

The matrix A given by A$\frac{1}{2}\left[\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right]$ is orthogonal matrix when$A{A}^T=1$ , whererepresent the identity matrix.

$$\begin{gathered} A{A}^T=\frac{1}{2}\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right)×\frac{1}{2}{\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right)}^T \\ =\left(\frac{1}{2}\right)\left(\frac{1}{2}\right){\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right)}_{2×2}{\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right)}_{2×2}^T \\ =\frac{1}{4}\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right){\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right)}^T \\ =\frac{1}{4}\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right)\left(\begin{array}{cc}-\sqrt{3}& 1\\ -1& -\sqrt{3}\end{array}\right) \end{gathered}$$

Further, solve.

$$\begin{gathered} A{A}^T=\frac{1}{4}\left(\begin{array}{cc}\left(-\sqrt{3}\right)\left(-\sqrt{3}\right)+\left(1\right)\left(1\right)& \left(-\sqrt{3}\right)(-1)+\left(1\right)\left(-\sqrt{3}\right)\\ (-1)\left(-\sqrt{3}\right)+\left(-\sqrt{3}\right)\left(1\right)& (-1)(-1)+\left(-\sqrt{3}\right)\left(-\sqrt{3}\right)\end{array}\right) \\ =\frac{1}{4}\left(\begin{array}{cc}3+1& \sqrt{3}-\sqrt{3}\\ \sqrt{3}-\sqrt{3}& 1+3\end{array}\right) \\ =\frac{1}{4}\left(\begin{array}{cc}4& 0\\ 0& 4\end{array}\right) \end{gathered}$$

Factor out 4 from the matrix.

$$\begin{gathered} A{A}^T=\left(\frac{1}{4}\right)\left(4\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\mathrm{l} \end{gathered}$$

Thus, it has been proved that A is an orthogonal matrix as$A{A}^T=\mathrm{l}$ , where represent the identity matrix.

The determinant of order two is determined by using the formula

$\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|=a_{11}{a}_{22}-a_{12}{a}_{21}$.

Find the determinant of A.

$$\begin{gathered} det\left(A\right)=\left|\begin{array}{cc}-\frac{\sqrt{3}}{2}& \frac{1}{2}\\ -\frac{1}{2}& -\frac{\sqrt{3}}{2}\end{array}\right| \\ =\left[\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\right] \\ =\left[\frac{3}{4}-\left(-\frac{1}{4}\right)\right] \\ =\frac{1}{4}-\frac{1}{4} \end{gathered}$$

Further, solve.

det (A) = 1

Any 2x2 orthogonal matrix with a determinant equal to 1 corresponds to a rotation, thus, it has to be shown that the following orthogonal matrix A is a rotation matrix.

Since the matrix A represents an active transformation of vectors in the ( x , y ) plane (axes fixed, vectors rotated, so we must use the following definition of the general form of the vector representation of an active rotation in two dimensions defined as

The vector $\stackrel{â‡\P Ä}{r}$from the origin to the point (x , y ) has been rotated by an angle to become the vector from the origin to the point ( X ,Y ) written in the matrix form$\left(\begin{array}{c}X\\ Y\end{array}\right)=\left(\begin{array}{cc}\cos \mathrm{θ}& -\sin \mathrm{θ}\\ \sin \mathrm{θ}& \cos \mathrm{θ}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$ called rotation equation which relates the components of and .

Compare this rotation matrix A with the general form of the matrix representation (rotation matrix) of an active rotation in two dimensions$\left(\begin{array}{cc}\cos \mathrm{θ}& -\sin \mathrm{θ}\\ \sin \mathrm{θ}& \cos \mathrm{θ}\end{array}\right)$ , it Is deduced that$\cos \left(\mathrm{θ}\right)=\frac{-\sqrt{3}}{2}$and$\sin \left(\mathrm{θ}\right)=-\frac{1}{2}.$ .

Find the rotation angle.

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\frac{-1/2}{-\sqrt{3}/2} \\ ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =\mathrm{ττ}+\frac{\mathrm{ττ}}{6} \\ =\frac{7\mathrm{ττ}}{6} \\ ={210}^{°} \end{gathered}$$

So, this is a rotation of${210}^{°}$ .