Answer
$A$ is an orthogonal matrix.
Work Step by Step
We have:
$|A|=\left|\begin{array}{ccc}
1/\sqrt 2 & 0 & -1/\sqrt 2 \\
0&1&0\\
1/\sqrt 2&0&1/\sqrt 2
\end{array}\right|=1$
The matrix $A$ is non-singular and $A$ is invertible.
The augmented matrix is
$ \left[\begin{array}{ccc}
1/\sqrt 2 & 0 & -1/\sqrt 2 \\
0&1&0\\
1/\sqrt 2&0&1/\sqrt 2
\end{array}\right]$
Thus:
$A^{-1}=\left[\begin{array}{ccc}
1/\sqrt 2 & 0 & 1/\sqrt 2 \\
0&1&0\\
-1/\sqrt 2&0&1/\sqrt 2
\end{array}\right]$
Note that:
$A^{T}=\left[\begin{array}{ccc}
1/\sqrt 2 & 0 & 1/\sqrt 2 \\
0&1&0\\
-1/\sqrt 2&0&1/\sqrt 2
\end{array}\right]$
Therefore, we see that
$A^{T}=A^{-1}$
Thus $A$ is an orthogonal matrix.