Answer
The matrix $A$ is orthogonal.
Work Step by Step
Let the matrix be
$A=\left[\begin{array}{cc}
1/\sqrt 2&-1/\sqrt 2\\
-1/\sqrt 2&-1/\sqrt 2
\end{array}\right]$
Then: $|A|=-1$
We have:
$A^{T}=\left[\begin{array}{cc}
1/\sqrt 2&-1/\sqrt 2\\
-1/\sqrt 2&-1/\sqrt 2
\end{array}\right]$
and $A^{-1}= (-1)* \left[\begin{array}{cc}
-1/\sqrt 2&1/\sqrt 2\\
1/\sqrt 2&1/\sqrt 2
\end{array}\right]=\left[\begin{array}{cc}
1/\sqrt 2&-1/\sqrt 2\\
-1/\sqrt 2&-1/\sqrt 2
\end{array}\right] = A^{T}$
Thus the matrix $A$ is orthogonal.
