Do the details of Example 3as follows:

(a) Verify that the four matrices in (7.14)are all orthogonal and verify the stated values of their determinants.

(b) Verify the products C=AB and D=BAin (7.15).

(c) Solve (7.16)to find the reflection line.

(d) Analyze the transformation Das we did C.

A square matrix is said to be orthogonal if ${\mathbf{A}}^{\mathbf{\tau }}\mathbf{A}\mathbf{=}\mathbf{l}\mathbf{=}{\mathbf{AA}}^{\mathbf{\tau }}$.

The given matrices are$\mathrm{A}=\frac{1}{2}\left(\begin{array}{cc}-1& \sqrt{3}\\ -\sqrt{3}& -1\end{array}\right),\mathrm{B}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\mathrm{C}=\mathrm{AB},\mathrm{D}=\mathrm{BA}$

In mathematics, matrix multiplication is a way that produces a matrix by multiplying two matrices.

$$\begin{gathered} C=\frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ -\sqrt{3}& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ -\sqrt{3}& 1\end{array}\right) \end{gathered}$$and

$$\begin{gathered} D=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\times \frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ -\sqrt{3}& -1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ -\sqrt{3}& 1\end{array}\right) \end{gathered}$$

As

$$\begin{gathered} {\mathrm{B}}^{\mathrm{\tau }}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=\mathrm{B} \\ {\mathrm{C}}^{\mathrm{\tau }}=\frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ -\sqrt{3}& 1\end{array}\right)=\mathrm{C} \\ {\mathrm{D}}^{\mathrm{\tau }}=\frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ \sqrt{3}& 1\end{array}\right)=\mathrm{D} \end{gathered}$$

And

role="math" localid="1658995917267" $$\begin{gathered} {\mathrm{AA}}^{\mathrm{\tau }}=\frac{1}{2}\left(\begin{array}{cc}-1& \sqrt{3}\\ -\sqrt{3}& -1\end{array}\right)\times \frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ \sqrt{3}& -1\end{array}\right) \\ =\frac{1}{4}\left(\begin{array}{cc}4& 0\\ 0& 4\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

Therefore all of these matrices are orthogonal.

The determinants of these matrices are

$$\begin{gathered} detA=\frac{1}{4}\left(\left(-1\right)\left(-1\right)-\sqrt{3}\left(-\sqrt{3}\right)\right)=1 \\ detC=\frac{1}{4}\left(\left(-1\right)\left(1\right)-\left(-\sqrt{3}\right)\left(-\sqrt{3}\right)\right)=-1 \\ detD=\frac{1}{4}\left(\left(-1\right)\left(1\right)-\left(\sqrt{3}\right)\left(\sqrt{3}\right)\right)=-1 \\ detB=1\left(-1\right)=-1 \end{gathered}$$

This means that A is a rotation and B,C and D are reflections

The vector r unchanged by C, that is solving the equation Cr=r which gives

$$\begin{gathered} \frac{1}{2}\left(\begin{array}{cc}-1& -\sqrt{3}\\ -\sqrt{3}& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \\ \frac{1}{2}\left(\begin{array}{c}-\mathrm{x}-\sqrt{3}\\ -\sqrt{3}+\mathrm{y}\end{array}\right)=\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \end{gathered}$$

We obtain two equations

$$\begin{gathered} -\frac{x}{2}-\frac{\sqrt{3}}{2}y=x, \\ -\frac{\sqrt{3}}{2}x+\frac{y}{2}=y, \end{gathered}$$

Which gives the same solution, .role="math" localid="1658996515324" $y=-\sqrt{3}x$.

Repeat the same analysis for the matrix D.

The reflection line is obtained as

$$\begin{gathered} \frac{1}{2}\left(\begin{array}{cc}-1& \sqrt{3}\\ \sqrt{3}& 1\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right)=\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \\ \frac{1}{2}\left(\begin{array}{c}-\mathrm{x}+\sqrt{3}\\ \sqrt{3}+\mathrm{y}\end{array}\right)=\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right) \end{gathered}$$

We obtain two equations

$$\begin{gathered} -\frac{x}{2}+\frac{\sqrt{3}}{2}y=x, \\ \frac{\sqrt{3}}{2}x+\frac{y}{2}=y, \end{gathered}$$

Which gives the same solution $y=\sqrt{3}x$.

This means that the vector$\mathrm{r}=\mathrm{i}+\sqrt{3}\mathrm{j}$is unchanged by the transformation. The vector perpendicular to$\mathrm{r}=\mathrm{i}+\sqrt{3}\mathrm{j}$saylocalid="1658996878466" $\mathrm{r}=\sqrt{3}-\mathrm{j}$is transformed as

$\frac{1}{2}\left(\begin{array}{cc}-1& \sqrt{3}\\ \sqrt{3}& 1\end{array}\right)\left(\left.\begin{array}{c}\sqrt{3}\\ -1\end{array}\right)=\left(\begin{array}{c}-\sqrt{3}\\ 1\end{array}\right.\right)$,

which is negative.

Therefore, all matrices are orthogonal, that$\mathrm{detA}=1$ and $\mathrm{detB}=\mathrm{detC}=\mathrm{detD}=-1$. The reflection line of C is$\mathrm{i}-\sqrt{3}\mathrm{j}$ and that the reflection line of D is $\mathrm{i}+\sqrt{3}\mathrm{j}$.