Question

Determine if the matrix p=
[√3/3, √6/6, -√2/2 ]
[-√3/3, √6/3, 0 ]
[√3/3, √6/6, √2/2] is othognal.

Answer 1

To determine, whether the matrix A is orthogonal we should multiply matrix to its transpose, if the result is equal to the identity matrix, then A is orthogonal:

$A=\begin {pmatrix} \cfrac{1}{\sqrt{3}} & \cfrac{1}{\sqrt{6}} & \cfrac{-1}{\sqrt{2}}\\ \cfrac{-1}{\sqrt{3}}& \cfrac{2}{\sqrt{6}}& 0\\ \cfrac{1}{\sqrt{3}} & \cfrac{1}{\sqrt{6}}& \cfrac{1}{\sqrt{2}} \end{pmatrix}$

$A^T=\begin{pmatrix} \cfrac{1}{\sqrt{3}} & \cfrac{-1}{\sqrt{3}} & \cfrac{1}{\sqrt{3}}\\ \cfrac{1}{\sqrt{6}}& \cfrac{2}{\sqrt{6}}& \cfrac{1}{\sqrt{6}}\\ \cfrac{-1}{\sqrt{2}} & 0& \cfrac{1}{\sqrt{2}} \end{pmatrix}$

$AA^T=$ $\begin{pmatrix} (\cfrac{1}{3}+\cfrac{1}{6}+\cfrac{1}{2}) & (\cfrac{-1}{3} + \cfrac{2}{6})& (\cfrac{1}{3}+\cfrac{1}{6}-\cfrac{1}{2})\\ (\cfrac{-1}{3} + \cfrac{2}{6})& (\cfrac{1}{3}+\cfrac{4}{6})& (\cfrac{-1}{3} + \cfrac{2}{6})\\ (\cfrac{1}{3}+\cfrac{1}{6}-\cfrac{1}{2}) & (\cfrac{-1}{3} + \cfrac{2}{6})& (\cfrac{1}{3}+\cfrac{1}{6}+\cfrac{1}{2}) \end{pmatrix}= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}$

So A is an orthogonal matrix.