Question

Show that matrix $A=\left[\begin{array}{ccc}\cos \theta & 0& \sin \theta \\ 0& 1& 0\\ -\sin \theta & 0& \cos \theta \end{array}\right]$ in orthogonal matrix.

Answer 1

Given, $A=\left[\begin{array}{ccc}\cos \theta & 0& \sin \theta \\ 0& 1& 0\\ -\sin \theta & 0& \cos \theta \end{array}\right]$.

Now,

$A.A^T=$$\left[\begin{array}{ccc}\cos \theta & 0& \sin \theta \\ 0& 1& 0\\ -\sin \theta & 0& \cos \theta \end{array}\right]$$\left[\begin{array}{ccc}\cos \theta & 0& -\sin \theta \\ 0& 1& 0\\ \sin \theta & 0& \cos \theta \end{array}\right]$

or, $A{A}^T=$ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$......(1).

Again,

$A^{T}A=$ $\left[\begin{array}{ccc}\cos \theta & 0& -\sin \theta \\ 0& 1& 0\\ \sin \theta & 0& \cos \theta \end{array}\right]$$\left[\begin{array}{ccc}\cos \theta & 0& \sin \theta \\ 0& 1& 0\\ -\sin \theta & 0& \cos \theta \end{array}\right]$

or, $A^{T}A=$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$.......(2).

From (1) and (2) we get,

$A{A}^T=A^{T}A=I_3$. [ Where $I_3$ is the identity matrix of order $3$].

So $A$ is an orthogonal matrix.