91Ó°ÊÓ

Let’s assume that${\stackrel{r}{v}}_1={\left(a,c\right)}^T,{\stackrel{r}{v}}_2={\left(b,d\right)}^T$ are the column vectors of an orthogonal matrix A.

Therefore, orthogonal conditions are$\left|\left|{\stackrel{âŠ\yen }{v}}_1\right|\right|=\left|\left|{\stackrel{âŠ\yen }{v}}_2\right|\right|=1$ and${\stackrel{âŠ\yen }{v}}_1.{\stackrel{âŠ\yen }{v}}_2=0$.

Let’s find the form of matrix.

$$\begin{gathered} \left|\left|{\stackrel{r}{v}}_1\right|\right|=1⇒\sqrt{a^2+c^2}=1 \\ ⇒a^2+c^2=1 \\ \left|\left|{\stackrel{r}{v}}_2\right|\right|=1⇒\sqrt{b^2+d^2}=1 \\ ⇒b^2+d^2=1 \\ {\stackrel{r}{v}}_1.{\stackrel{r}{v}}_2=0⇒\left[ac\right]\left[\begin{array}{c}b\\ d\end{array}\right] \\ ⇒ab+cd=0 \end{gathered}$$

Now use these equations to make conclusion about the entries of A.

$\begin{array}{r}{a}^2+c^2=1âˆ\S b^2+d^2=1âˆ\S ab+cd=0\\ ⇒a^2=1−c^2âˆ\S b^2=1−d^2âˆ\S ab=−cd\\ ⇒a^2=1−c^2âˆ\S d^2=1−b^2âˆ\S (ab{)}^2=(−cd{)}^2\\ ⇒a^2=1−c^2âˆ\S d^2=1−b^2âˆ\S a^2{b}^2=c^2{d}^2\\ ⇒a^2=1−c^2âˆ\S d^2=1−b^2âˆ\S \left(1−c^2\right)b=c^2\left(1−b^2\right)\\ ⇒a^2=1−c^2âˆ\S d^2=1−b^2âˆ\S b^2−c^2{b}^2=c^2−c^2{b}^2\\ ⇒a^2=1−c^2âˆ\S d^2=1−b^2âˆ\S b^2=c^2\\ ⇒a^2=1−b^2âˆ\S d^2=1−b^2âˆ\S b^2=c^2\\ ⇒a^2=d^2âˆ\S b^2=c^2âˆ\S ab+cd=0\\ ⇒d=Â\pm aâˆ\S b^2=c^2âˆ\S ab+cd=0\\ ⇒d=Â\pm aâˆ\S b^2=c^2âˆ\S ab+c(Â\pm a)=0\\ ⇒d=Â\pm aâˆ\S b^2=c^2âˆ\S a(b+Â\pm c)=0\\ ⇒d=Â\pm aâˆ\S b^2=c^2âˆ\S (b+Â\pm c)=0,(a≠0)\\ ⇒d=Â\pm aâˆ\S c=\mathrm{mb}\end{array}$

Hence, the solution is any $2×2$orthogonal matrix has one of the following forms is

$\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]v\left[\begin{array}{cc}a& b\\ b& -a\end{array}\right],a,b∈R,a^2+b^2=1$.