91Ó°ÊÓ

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 be said that, 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 $n×n$order and $A^T$ is the transpose of A. Then according to the definition, if, $A^T=A^{-1}$is satisfied, then,

$A·A^T=I$

For example, take

$$\begin{gathered} A=\left[cos\mathrm{θ}-sin\mathrm{θ} \\ sin\mathrm{θ}cos\mathrm{θ}\right] \end{gathered}$$and $$\begin{gathered} B=\left[cos\mathrm{θ}sin\mathrm{θ} \\ -sin\mathrm{θ}cos\mathrm{θ}\right] \end{gathered}$$

Clearly, A and B are orthogonal; matrices, now observe that

$$\begin{gathered} A+B=\left[2cos\mathrm{θ}0 \\ 02cos\mathrm{θ}\right] \end{gathered}$$

Therefore, if we choose $\mathrm{θ}∈\left[0,2\mathrm{π}\right]$then $A+B$will be orthogonal.

So, take $\mathrm{θ}=\frac{\mathrm{π}}{3}$

Then,

$$\begin{gathered} A=\left[\frac{1}{2}\frac{-\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2}\frac{1}{2}\right] \\ B=\left[\frac{1}{2}\frac{\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{2}\frac{1}{2}\right] \end{gathered}$$

Hence, the given statement is true.