91Ó°ÊÓ

There are two first rows of two individual orthogonal matrices. One is from an \(2 \times 2\) orthogonal matrix, and the other one is from an \(3 \times 3\) orthogonal matrix.

Let us consider the second row is \(\left( {\begin{align}{}x&y\end{align}} \right)\) .

So, the matrix becomes \(A = \left( {\begin{align}{}{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\x&y\end{align}} \right)\)

This is an orthogonal matrix. So, this can be concluded that,

\(\begin{array}{l}\left[ {\begin{array}{}{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\x&y\end{array}} \right] \times \left[ {\begin{array}{}{\frac{1}{{\sqrt 2 }}}&x\\{\frac{1}{{\sqrt 2 }}}&y\end{array}} \right] = \left[ {\begin{array}{}1&0\\0&1\end{array}} \right]\\ \Rightarrow \left[ {\begin{array}{}1&{\frac{x}{{\sqrt 2 }} + \frac{y}{{\sqrt 2 }}}\\{\frac{x}{{\sqrt 2 }} + \frac{y}{{\sqrt 2 }}}&{{x^2} + {y^2}}\end{array}} \right] = \left[ {\begin{array}{}1&0\\0&1\end{array}} \right]\end{array}\)

Thus,

\(\begin{align}\frac{x}{{\sqrt 2 }} + \frac{y}{{\sqrt 2 }} &= 0 \cdots \left( 1 \right)\\{x^2} + {y^2} &= 1 \cdots \left( 2 \right)\end{align}\)

So, from (1) and (2), there can get that,

\(y = \frac{1}{{\sqrt 2 }}\;and\;x = - \frac{1}{{\sqrt 2 }}\)

Therefore, the orthogonal matrix is \(\left( {\begin{align}{}{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\\{ - \frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}\end{align}} \right)\).

Here the first row of the orthogonal matrix is \(\left( {\begin{align}{}{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\end{align}} \right)\). That is this is a unit vector.

Now, consider a perpendicular vector and normalize it. So, assume it is \(\left( {\begin{align}{}{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 2 }}}&0\end{align}} \right)\) .

it is the second row of the orthogonal matrix.

Therefore, a vector perpendicular to the first two vectors and a unit length. The cross-product will help to produce such a vector.

So, the third row is \(\left( {\begin{align}{}{\frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 6 }}}&{ - \frac{2}{{\sqrt 6 }}}\end{align}} \right)\). The dot product with the other two vectors is 0 and has a unit length.

Thus, the required orthogonal matrix is \(\left( {\begin{align}{}{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 2 }}}&0\\{\frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 6 }}}&{ - \frac{2}{{\sqrt 6 }}}\end{align}} \right)\) .