Answer to Question #94878 in Linear Algebra for Chief

Question

Determine if the matrix p=
D= 1/√2, 0, 1/√2
0, 1, 0
-1/√2, 2, 1/√2
Is othognal.

Answer 1

This matrix would be orthogonal if a product of the matrix and its transpose is a unit matrix:

DD^T = E

Thus, we should first found a transpose matrix:

$D= \begin{vmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{-1}{\sqrt{2}} & 2 & \frac{1}{\sqrt{2}} \end{vmatrix} D^T= \begin{vmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{-1}{\sqrt{2}} \\ 0 & 1 & 2 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{vmatrix}$

Now, we must multiply them together:

$DD^T= \begin{vmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{-1}{\sqrt{2}} & 2 & \frac{1}{\sqrt{2}} \end{vmatrix} \begin{vmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{-1}{\sqrt{2}} \\ 0 & 1 & 2 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 5 \end{vmatrix}$

The product is not the unit matrix. Thus, the D matrix was not orthogonal.