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A square matrix with real numbers or elements is said to be an orthogonal matrix, if its transpose is equal to its inverse matrix.

Or it can say, when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as an orthogonal matrix.

For example,

Suppose A is a square matrix with real elements and of n x norder and A^Tis the transpose of A. Then according to the definition, if, A^T = A^-1 is satisfied, then,

A A^T = I

It wants to find all orthogonal matrices of the form,

$\left[\begin{array}{ccc}a& b& 0\\ c& d& 1\\ e& f& 0\end{array}\right]$

Let ${\stackrel{→}{r}}_i$ be the rows and ${\stackrel{→}{c}}_i$be the columns of the matrix A. now observe,

$\left[\begin{array}{ccc}a& b& 0\\ c& d& 1\\ e& f& 0\end{array}\right]\underset{{\stackrel{→}{c}}_1.{\stackrel{→}{c}}_3=0⇒d=0}{\overset{{\stackrel{→}{c}}_2.{\stackrel{→}{c}}_3=0⇒d=0}{→}}\left[\begin{array}{ccc}a& b& 0\\ 0& 0& 1\\ e& f& 0\end{array}\right]$

The last matrix will be orthogonal if and only if the matrix $\left[\begin{array}{cc}a& b\\ e& f\end{array}\right]$is orthogonal.

To check that the above matrix is orthogonal or not consider the solution below.

Observe that,

$$\begin{gathered} \left[\begin{array}{ccc}a& b& 0\\ 0& 0& 1\\ e& f& 0\end{array}\right].\left[\begin{array}{ccc}a& 0& e\\ b& 0& f\\ 0& 1& 0\end{array}\right]=\left[\begin{array}{ccc}{a}^2+b^2& 0& ae+bf\\ 0& 1& \\ ae+bf& 0& e^2+f^2\end{array}\right] \\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ ⇔\left[\begin{array}{cc}{a}^2+b^2& ae+bf\\ ae+bf& e^2+f^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ ⇔\left[\begin{array}{cccc}a& b& a& e\\ e& f& b& f\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Thus, it only has to see when does the matrix $\left[\begin{array}{cc}a& b\\ e& f\end{array}\right]$ is orthogonal.

But from Exercise 5.3.34 $2×2$matrix is orthogonal if and only if it is one of the following.

$\left[\begin{array}{ccc}\cos \mathrm{θ}& -\sin \mathrm{θ}& 0\\ 0& 0& 1\\ \sin \mathrm{θ}& \cos \mathrm{θ}& 0\end{array}\right]or\left[\begin{array}{ccc}\cos \mathrm{θ}& \sin \mathrm{θ}& 0\\ 0& 0& 1\\ \sin \mathrm{θ}& -\cos \mathrm{θ}& 0\end{array}\right]$

Hence, the matrix A is orthogonal if it will be in the form.

$\left[\begin{array}{ccc}\cos \mathrm{θ}& -\sin \mathrm{θ}& 0\\ 0& 0& 1\\ \sin \mathrm{θ}& \cos \mathrm{θ}& 0\end{array}\right]or\left[\begin{array}{ccc}\cos \mathrm{θ}& \sin \mathrm{θ}& 0\\ 0& 0& 1\\ \sin \mathrm{θ}& -\cos \mathrm{θ}& 0\end{array}\right]$, $\mathrm{θ}∈[0,2\mathrm{π})$.