Show that each of the following matrices is orthogonal and find the rotation and/or reflection it produces as an operator acting on vectors. If a rotation, find the axis and angle; if a reflection, find the reflecting plane and the rotation, if any, about the normal to that plane.

$\mathit{M}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\left(\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right)$

The given matrix is$M=\frac{1}{3}\left(\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right)$

The inverse of a matrix is mathematically presented as${\mathbf{\text{M}}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{d}\mathbf{e}\mathbf{t}\mathbf{\text{M}}}{\mathit{C}}^{\mathbf{\text{T}}}$

Calculate the inverse to verify that the matrix is orthogonal.

${\text{M}}^{-1}=\frac{1}{\det \text{M}}{C}^{\text{T}}$

The determinant is as shown below.

$$\begin{gathered} M=\frac{1}{3^3}(-1+8+8-(-4)-(-4)-(-4\left)\right) \\ =\frac{27}{27} \\ =1 \end{gathered}$$

The matrix of cofactors is $C=\frac{1}{3}\left(\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right)$. This gives the inverse

$$\begin{gathered} M^{-1}=\frac{1}{3}\left(\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right) \\ =M^T \\ =M \end{gathered}$$

Use the property $M=M^{-1}$ because the matrix is symmetric. Since its determinant is 1, it is a rotation.

To find the axis of rotation solve the equation $Mr=r$, i.e., find the eigenvector that corresponds to the eigenvalue 1.

$$\begin{gathered} \frac{1}{3}\left(\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \frac{1}{3}\left(\begin{array}{ccc}-4& 2& 2\\ 2& -4& 2\\ 2& 2& -4\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=0 \end{gathered}$$

The above equation gives the equations $-4x+2y+2z=0,2x-4y+2z=0,2x+2y-4z=0$. Subtract the second from the first equation to get $x=y$. Subtract the last from the first equation and get $x=z$. Then the eigenvector is $r=i+j+k$.

To find the angle of rotation, note that $M^2=1$, which means that the angle of rotation is $180°$. Alternatively, diagonalize matrix $M$by solving the equation as shown below.

$\left|\begin{array}{ccc}-1-3\lambda & 2& 2\\ 2& -1-3\lambda & 2\\ 2& 2& -1-3\lambda \end{array}\right|=0$

$$\begin{gathered} -{(1+3\lambda )}^3+16+12(1+3\lambda )=0 \\ {\lambda }^3-{\lambda }^2-\lambda -1=0 \\ (\lambda +1)({\lambda }^2-1)=0 \end{gathered}$$

The above equation gives the eigenvalues $-1,-1,1$

The diagonal formed is $D=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right)$. As expected, this gives $\cos \theta =-1$, that is, $\theta =\pi$.