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.

$\mathit{M}\mathbf{=}\frac{\mathbf{1}}{\mathbf{9}}\left(\begin{array}{ccc}-1& 4& 4\\ -4& -4& 7\\ -8& 1& -4\end{array}\right)$

The given matrix is$M=\frac{1}{9}\left(\begin{array}{ccc}-1& 8& 4\\ -4& -4& 7\\ -8& 1& -4\end{array}\right)$

The inverse of a matrix can be 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 its inverse to verify that the matrix is orthogonal.

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

The determinant is equal to $\det M=\frac{1}{9^3}(-16-448-16-128-128+7)=-1$

The matrix of cofactor is $C=\frac{1}{9}\left(\begin{array}{ccc}1& -8& -4\\ 4& 4& -7\\ 8& -1& 4\end{array}\right)$

This gives the inverse $M^{-1}=\frac{1}{9}\left(\begin{array}{ccc}-1& -4& -8\\ 8& -4& 1\\ 4& 7& -4\end{array}\right)=M^T$

So, the matrix is orthogonal. Since its determinant is $-1$, it is a reflection.

To find the plane of reflection, solve the eigenvalue equation $\mathit{M}\mathit{r}\mathbf{=}\mathbf{-}\mathit{r}$

$$\begin{gathered} \frac{1}{9}\left(\begin{array}{ccc}-1& 8& 4\\ -4& -4& 7\\ -8& 1& -4\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=-\left(\begin{array}{c}x\\ y\\ z\end{array}\right) \\ \frac{1}{9}\left(\begin{array}{ccc}8& 8& 4\\ -4& 5& 7\\ -8& 1& 5\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=0 \end{gathered}$$

This

The above equation gives the equations $8x+8y+4z=0,-4x+5y+7z=0,-8x+y+5z=0$. Add the second and third equations to obtain $-2x+y+2z=0$. By adding this equation and the first, we obtain $y=-z$. By inserting this into the first equation, we get $x=z/2$. Then the vector is $\mathbf{r}=\mathbf{i}-2\mathbf{j}+2\mathbf{z}$, which gives the plane of reflection $x-2y+2z=0.$

Other two eigenvalues by solving

$$\begin{gathered} \left|\begin{array}{ccc}-1-9\lambda & 8& 4\\ -4& -4-9\lambda & 7\\ -8& 1& -4-9\lambda \end{array}\right|=0 \\ -(1+9\lambda ){(4+9\lambda )}^2-464-64(4+9\lambda )+7(1+9\lambda )=0 \\ {\lambda }^3+{\lambda }^3+\lambda +1=0 \\ ({\lambda }^2+1)(\lambda +1)=0 \end{gathered}$$

The above equation gives the eigenvalues $-1,i$, $-i$. Since the trace of a reflection is $2\cos \theta -1$, and the sum of the eigenvalues is $-1$, $\cos \theta =0$, it is $90°$ again.