Question

TRUE OR FALSE

There exist orthogonal 2×2 matrices Aand Bsuch that A+Bis orthogonal as well.

Short answer

The given statement is true.

Step-by-step solution

Definition of an orthogonal matrix

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.

Consider an example

For example,

Suppose A is a square matrix with real elements and of $n\times n$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

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

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

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

Therefore, if we choose $\theta \in \left[0,2\pi \right]$then $A+B$will be orthogonal.

So, take $\theta =\frac{\pi }{3}$

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.