91Ó°ÊÓ

Inlinear algebra, an orthogonal matrix, is a realsquare matrixwhose columns and rows areorthonormalvectors.

One way to express this is

${\mathit{Q}}^{\mathbf{T}}\mathit{Q}\mathbf{=}\mathit{Q}{\mathit{Q}}^{\mathbf{T}}\mathbf{=}\mathit{I}$

Where${\mathit{Q}}^{\mathbf{T}}$is thetransposeof QandIis theidentity matrix.

For example,

Consider the matrix below.

$10A=\left[\begin{array}{l}ab\\ cd\end{array}\right]$

With a, b, c and d integers.

$A=\left[\begin{array}{l}a/10b/10\\ c/10d/10\end{array}\right]$

Must be an orthogonal matrix, so

$\begin{array}{c}{\left(\frac{a}{10}\right)}^2+{\left(\frac{c}{10}\right)}^2=1\\ a^2+c^2=1\end{array}$

There are only two ways to write 100 as a sun of two positive perfect squares 100=36+64=64+36.

Since, a and c are required to be positive, we have either a=6 and c=8 or a=8 and c=6.

In the each case mentioned above there are only two options for the second column of A,

The two unit vectors perpendicular to the first column vector.

Four solutions:

$A=\left[\begin{array}{l}0.6-0.8\\ 0.80.6\end{array}\right],\left[\begin{array}{l}0.60.8\\ 0.8-0.6\end{array}\right],\left[\begin{array}{l}0.8-0.6\\ 0.60.8\end{array}\right]$

Thus,$A=\left[\begin{array}{l}0.80.6\\ 0.6-0.8\end{array}\right]$

Hence, the solution for the matrix A is $A=\left[\begin{array}{l}0.80.6\\ 0.6-0.8\end{array}\right]$.