Question

$A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$is an orthogonal matrix

• A. True
• B. False

Answer 1

The correct option is A

True

We know that the condition for a matrix A to be called as a orthogonal matrix is

$A\times A^T=A^T\times A=I$ Lets calculate $A\times A^T$ first

$A^T=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]SoA{A}^T=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$
$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=1$
So A $A^T=I$
Similarly A $A^T\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$
$=A^T\times A$
$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$

So $A\times A^T=A^T\times A=I$. Hence A is an orthogonal matrix and the statement is true.

Also here note that A $A^T$ is not necessarily equal to $A^{T}A$ because matrix multiplication is not commutative. We should always check both the conditions separately.