# How to determine a and b so that the real matrix ((a,1,b),(b,a,1),(1,b,a)) is orthogonal matrix ?

Apr 9, 2017

$a = b = 0$

#### Explanation:

A matrix $M$ is orthogonal if and only if:

$M {M}^{T} = {M}^{T} M = I$

Evaluating this directly with the given form of matrix:

$\left(\begin{matrix}a & 1 & b \\ b & a & 1 \\ 1 & b & a\end{matrix}\right) {\left(\begin{matrix}a & 1 & b \\ b & a & 1 \\ 1 & b & a\end{matrix}\right)}^{T}$

$= \left(\begin{matrix}a & 1 & b \\ b & a & 1 \\ 1 & b & a\end{matrix}\right) \left(\begin{matrix}a & b & 1 \\ 1 & a & b \\ b & 1 & a\end{matrix}\right)$

$= \left(\begin{matrix}{a}^{2} + {b}^{2} + 1 & a b + a + b & a b + a + b \\ a b + a + b & {a}^{2} + {b}^{2} + 1 & a b + a + b \\ a b + a + b & a b + a + b & {a}^{2} + {b}^{2} + 1\end{matrix}\right)$

In order for this to be the identity matrix, we require:

${a}^{2} + {b}^{2} + 1 = 1$

So if $a$, $b$ are real then they must both be $0$.

Then we find:

$a b + a + b = 0$

satisfying the requirement that the off diagonal elements be $0$.

Apr 9, 2017

See below

#### Explanation:

Also, an orthogonal matrix has columns and rows that are orthogonal unit vectors.

For the first column vector:

$\left\mid m a t h b f {v}_{1} \right\mid = \sqrt{{a}^{2} + {b}^{2} + {1}^{2}} = 1 \implies a = b = 0 \text{ if } a , b \in \mathbb{R}$.