91Ó°ÊÓ

The given matrix is$\frac{1}{2}\left(\begin{array}{ccc}1& \sqrt{2}& -1\\ \sqrt{2}& 0& \sqrt{2}\\ 1& -\sqrt{2}& -1\end{array}\right)$

The inverse of a matrix is given by

${\mathbf{M}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{detM}}{\mathbf{C}}^{\mathbf{T}}$

Calculate its inverse to verify that the matrix is orthogonal${\mathrm{M}}^{-1}=\frac{1}{\mathrm{detM}}{\mathrm{C}}^{\mathrm{T}}$

The determinant is given as

$$\begin{gathered} detM=\frac{1}{2^3}\left(0+2-2-0\left(-2\right)-\left(2\right)\right) \\ =\frac{8}{8} \\ =1 \end{gathered}$$

The matrix of cofactors is given as

role="math" localid="1658819628728" $C=\frac{1}{2}\left(\begin{array}{ccc}1& \sqrt{2}& -1\\ \sqrt{2}& 0& \sqrt{2}\\ 1& -\sqrt{2}& -1\end{array}\right)$

which gives the inverse

role="math" localid="1658819808945" $$\begin{gathered} {\mathbf{M}}^{\mathbf{-}\mathbf{1}}=\frac{1}{2}\left(\begin{array}{ccc}1& \sqrt{2}& -1\\ \sqrt{2}& 0& \sqrt{2}\\ 1& -\sqrt{2}& -1\end{array}\right) \\ ={\mathrm{M}}^T \end{gathered}$$

Since its determinant is 1.

To find the axis of rotation, solve the equation $Mr=\mathit{r}$, that is, find the eigenvector corresponds to the eigenvalue 1.

This gives

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

This gives the equations

$-x+\sqrt{2}y-z=0, \sqrt{2}x-2y+\sqrt{2}z=0, x-\sqrt{2}y-3z=0$

Subtract the second from the first equation, get x=y . add the first and the last equations, get z=0. The first equation then gives$x=\sqrt{2}y$ .

The eigenvector is then

$\mathit{r}=\sqrt{2}\mathbf{}\mathit{i}+\mathit{j}$

To find the angle of rotation, diagonalize the matrix, solve the equation

$$\begin{gathered} \left|\begin{array}{ccc}1-2\mathrm{λ}& \sqrt{2}& -1\\ \sqrt{2}& -2\mathrm{λ}& \sqrt{2}\\ 1& -\sqrt{2}& -1-2\mathrm{λ}\end{array}\right|=0 \\ (1-2\mathrm{λ})(1+2\mathrm{λ})\left(2\mathrm{λ}\right)+4-2\mathrm{λ}+2(1+2\mathrm{λ})+2(1-2\mathrm{λ})=0 \\ (1-4{\mathrm{λ}}^2)\left(2\mathrm{λ}\right)+8-2\mathrm{λ}=0 \end{gathered}$$ ,

it is a rotation.

${\mathrm{λ}}^3=1$

This gives the eigenvalues $e^{2n\mathrm{Ï€¾\pm }/3}$, where n=0,1,2.

This gives ${\mathrm{λ}}_1=1,{\mathrm{λ}}_2=e^{i2\mathrm{π}/3}=-1/2+i\sqrt{3}/2$and ${\mathrm{λ}}_3=e^{i4\mathrm{π}/3)}=-1/2-i\sqrt{3}/2$. use that the sum of eigenvalues for a rotation is equal to $2\cos \mathrm{θ}+1$. The sum of eigenvalues is zero, which gives $\cos \mathrm{θ}=-1/2,$that is,$\mathrm{θ}=2\mathrm{π}/3$, which is $120°$. This could have also been concluded by noting that $M^3=1$.