91Ó°ÊÓ

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

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

If the determinant of the square matrix is equal to 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

In order to check the orthogonality of the matrix,${\mathrm{CC}}^{\mathrm{T}}=\mathrm{l}$condition must be satisfied.

To obtain the transpose of the matrix$\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$ , write the first row of the matrix, ( 0 ,-1 ) as the first column, and the second row of the matrix, ( -1 , 0 ) as the second column.

Obtain the transpose of the matrix$C^T$ .

$$\begin{gathered} C=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right] \\ {C}^T=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right] \end{gathered}$$

Multiply the$C^T$ with C .

$$\begin{gathered} C{C}^T=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right] \\ =\left[\begin{array}{cc}(0×0)+(-1×-1)& (0×-1)+(-1×0)\\ (-1×0)+(0×-1)& (-1×-1)+(0×0)\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

The product of $C{C}^T$is equal to the identity matrix l , which is$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ , the matrix is orthogonal.

The determinant (C) of the given square matrix $\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$is obtained by using the formula, $det\left(C\right)=\left(a_{11}×a_{22}\right)-\left(a_{12}×a_{21}\right)$.

Substitute ,, and in and obtain the value of

.

$$\begin{gathered} det\left(C\right)=\left(a_{11}×a_{22}\right)-\left(a_{12}×a_{21}\right) \\ =(0×0)-(-1×-1) \\ =-1 \end{gathered}$$

The determinant of the matrix det (C) is -1 .

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

The square matrix represents an active transformation of the vectors in (x,y) plane vector reflected through an det (C) = -1 .

To find the line of reflection, through the plane (x, y) , assume that the vectors along the plane are unchanged, find the value of x and y , the vector $\stackrel{â‡\P Ä}{r}$remains unchanged.

The orthogonal matrix C is given as $\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$.

The equivalent matrix form is $Cr=r\%$, substitute C as $\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$and r as

$(x,y)\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right].$

The matrix $\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]$is taken as $M_{22}$with 2 rows and 2 columns, and the matrix $\left[\begin{array}{c}x\\ y\end{array}\right]$is taken as $M_{21}$. The product of$M_{22}$ and $M_{21}$is obtained by the matrix of both$M_{22}$ and$M_{21}$ .

$$\begin{gathered} M_{21}×M_{22}=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right]×\left[\begin{array}{c}x\\ y\end{array}\right] \\ =\left[\begin{array}{c}(0×x)+(-1×y)\\ (-1×y)+(0×x)\end{array}\right] \\ =\left[\begin{array}{c}-y\\ -x\end{array}\right] \end{gathered}$$

Substitute in the equation is the value of the$M_{21}×M_{21}$ as$\left[\begin{array}{c}-y\\ -x\end{array}\right]$ in the equation.

$\left[\begin{array}{c}-y\\ -x\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]$

From the above equation, the value of x is -y and y is -x .

The vectors along the line of reflection have not changed, the x = -y is the line of reflection and the matrix is orthogonal.