# Find all possible values of $p$, $q$ and $r$ such that this matrix is orthogonal

Find all possible values of $p$, $q$ and $r$ such that the following matrix is orthogonal.

$$B= \begin{pmatrix} \frac1{3}&\frac2{3}&\frac2{3} \\ \frac2{3}&\frac1{3}&-\frac2{3} \\ p&q&r \end{pmatrix}$$

• Please do not images, but use mathias for the matrix. – Dietrich Burde Mar 10 '18 at 17:25
• That’s “MathJax” for formatting. Gotta love auto-correct. – amd Mar 10 '18 at 20:35

If $B$ is your matrix, then$$B^T.B=\begin{pmatrix}p^2+\frac{5}{9} & p q+\frac{4}{9} & p r-\frac{2}{9} \\ p q+\frac{4}{9} & q^2+\frac{5}{9} & q r+\frac{2}{9} \\ p r-\frac{2}{9} & q r+\frac{2}{9} & r^2+\frac{8}{9}\end{pmatrix}$$and you want this to be the identity matrix. So, since you only want to have $1$'s in the main diagonal, $p=q=\pm\frac23$ and $r=\pm\frac13$. Can you take it from here?

• Categorically accurate and didactic at the same time. – Piquito Mar 10 '18 at 19:21

$(p,q,r)$ is either the cross product of the first two rows, or its negative.

Hint: Compute $BB^T-I_3$. This must be zero; this way you obtain the equations

$$\begin{array}{rl}p+2q+2r &= 0\\ 2p+q-2r &= 0\\ p^2+q^2+r^2 &= 1\end{array}$$

This is easy to solve.

All columns of an orthogonal matrix are unit vectors. This give you two possible choices for each of $p,q,r$. Then you may test which combinations of choices are correct by checking whether the bottom row is orthogonal to the first two rows.