Question

Find an orthogonal matrix of the form $\left[\begin{array}{ccc}2/3& 1/\sqrt{2}& a\\ 2/3& -1/\sqrt{2}& b\\ 1/3& 0& c\end{array}\right]$.

Short answer

The orthogonal matrix for is $\left[\begin{array}{ccc}2/3& 1/\sqrt{2}& a\\ 2/3& -1/\sqrt{2}& b\\ 1/3& 0& c\end{array}\right]\mathrm{is}\left[\begin{array}{c}a=-1/\sqrt{18}\\ b=-1/\sqrt{18}\\ c=4/\sqrt{18}\end{array}\right]\mathrm{and}\left[\begin{array}{c}\mathrm{a}=1/\sqrt{18}\\ \mathrm{b}=1/\sqrt{18}\\ \mathrm{c}=-4/\sqrt{18}\end{array}\right]$.

Step-by-step solution

Determine the definition of an orthogonal matrix

A square matrix with real numbers or elements is said to be an orthogonal matrix, if its transpose is equal to its inverse matrix.

In other words, can say, when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as an orthogonal matrix.

For example,

Suppose A is a square matrix with real elements and of n x n order and A^Tis the transpose of A. Then according to the definition, if A^T = A^-1 is satisfied, then,

A A^T = I.

Determine the cross products

By definition of orthogonal matrix, a possible solution is to take cross product of $v_1,v_2$to obtain $v_3$

So, the cross product of is below localid="1660107116805" $v_1,v_2$.

$$\begin{gathered} v_1=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right] \\ {v}_2=\left[\begin{array}{c}1/\sqrt{2}\\ -1\sqrt{2}\\ 0\end{array}\right] \\ \mathrm{Now},\mathrm{the}\mathrm{cross}\mathrm{product}\mathrm{of}\mathrm{them}, \\ {\mathrm{v}}_1\mathrm{x}{\mathrm{v}}_2=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]\times \left[\begin{array}{c}1/\sqrt{2}\\ -1\sqrt{2}\\ 0\end{array}\right] \\ =\left[\begin{array}{c}1/\sqrt{18}\\ 1/\sqrt{18}\\ -4/\sqrt{18}\end{array}\right] \end{gathered}$$

Thus, the values of a, b and c is $1/\sqrt{18},1/-\sqrt{18},\mathrm{and}-4/\sqrt{18}$respectively.

Cross product of $v_2\mathrm{and}{v}_1\mathrm{is}{v}_2\times v_1=\left[\begin{array}{c}-1/\sqrt{18}\\ -1/\sqrt{18}\\ 4/\sqrt{18}\end{array}\right]$.

$$\begin{gathered} \mathrm{Thus},\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{is},-1/\sqrt{18},-1/\sqrt{18}\mathrm{and}4/\sqrt{18}\mathrm{respectively}. \\ \mathrm{Hence},\mathrm{the}\mathrm{orthogonal}\mathrm{matrix}\mathrm{is}\left[\begin{array}{c}\mathrm{a}=-1/\sqrt{18}\\ \mathrm{b}=-1/\sqrt{18}\\ \mathrm{c}=4/\sqrt{18}\end{array}\right]\mathrm{and}\left[\begin{array}{c}a=1/\sqrt{18}\\ b=1/\sqrt{18}\\ c=-4/\sqrt{18}\end{array}\right]. \end{gathered}$$