Answer to Question #169345 in Linear Algebra for Mohammad Hossain

Question

Determine whether the matrix (

1/√3 1/√6 −1/√2

1/√3 −2/√6 0

1/√3 1/√6 1/√2

) is Orthogonal or not.

Answer 1

Solution.

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

An orthogonal matrix is ​​a nondegenerate square matrix $A$ (usually with real elements) such that

$AA^{T}=A^{T}A=I,$ where $A^{T}$ is transposed matrix to matrix $A$, and $I$ is the identity matrix.

$\det A=-\frac{1}{3}-\frac{1}{6}-\frac{1}{3}-\frac{1}{6}=-1\neq0.$

$$A^{T}=\begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \end{pmatrix}$$ $$AA^{T}=\begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \end{pmatrix}=\newline =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}=I.$$ $$A^{T}A=I.$$

Answer. Yes, $A$ is orthogonal matrix.