I am given this problem. I need to find all $a,b,c,d \in \mathbb{R}$ for which the matrix $A$ is orthogonal.
$$A:=\frac{1}{3} \begin{pmatrix} -1 & 2 & a \\ 2 & 2 & b \\ 2 & c & d \end{pmatrix}$$
I know that $A$ is orthogonal if $A^T=A^{-1}$. But I am stuck how to find all unknowns and prove its orthogonality.
Can you guys please help me out?
On
$A^T=A^{-1}$ can also be written as $2$ equations given by $AA^{T}=I$ and $A^{T}A=I$ where $I$ is the identity matrix. Substitute for $A$ and you will get values of all variables. Hint: Try to solve for variable $c$ initially. That should reduce the whole thing to a set of under determined linear equations. Don't forget to scale the vectors.
Let $u,v,w$ be the first, second and third collumns of $3A$, respectively.
You wish to have $u\cdot v=0\wedge u\cdot w=0 \wedge v\cdot w=0$ (among other things).
This yields $-2+4+2c=0 \wedge -a+2b+2d= 0\wedge 2a+2b+cd=0$. Note that this is a linear system in (a very poor) disguise. So: $$\begin{align} \begin{cases} -2+4+2c=0 \\ -a+2b+2d= 0 \\ 2a+2b+cd=0 \end{cases}&\iff \begin{cases}c=-1\\-a+2b+2d= 0 \\2a+2b-d=0 \end{cases}\\ &\iff \begin{cases}c=-1\\ -a+2b+2d= 0\\ 3a+0b-3d=0 \end{cases}\\ &\iff \begin{cases}c=-1 \\a+2b=0\\a=d\end{cases}\\ &\iff\begin{cases}c=-1\\ b=-a/2\\ a=d\end{cases} \end{align}$$
and you get $3A=\begin{pmatrix} -1 & 2 & a \\ 2 & 2 & -a/2 \\ 2 & -1 & a \end{pmatrix}$
Now you want $||(a,-a/2,a)||=3$, that is $\displaystyle \sqrt {a^2+\frac{a^2}{4} +a^2}=3$, i.e., $\displaystyle \sqrt \frac{9a^2}{4}=3$ which gives $a=\pm 2.$