Given question is wrong.
if A is an orthogonal matrix then $det A=\pm 1$
Proof:
$\because AA^t=I$
$\Rightarrow |A A^t|=|I|$
$\Rightarrow |A| |A^t|=1$
$\Rightarrow|A|^2=1$
$\Rightarrow |A|=\pm 1$
But in the given question-
$det \; A= 1*2*4 (\text{product of eigen values}) =8 \neq \pm 1$
So, the given matrix is not an orthogonal matrix.
Matrix $A$ and $A^t$ have the same eigen values.
In case of orthogonal matrix, $AA^{t} = I$
Pre-multiplying by $A^{-1}$ both sides,
we get, $A^{t}=A^{-1}$
$\implies $eigen values of $A$ = eigen values of $A^t$= eigen values of $A^{-1}$
So the Trace of these matrices should also be equal in case of orthogonal matrices.
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Proof $1$ :- Matrices $A$ and $A^{t}$ have the same eigen values.
If $\lambda$ is an eigen value of a square matrix $A$
then $|A-\lambda I| = 0$
$\Rightarrow |(A-\lambda I)^{t}|=0$
$\Rightarrow |(A^t-\lambda I)|=0$
So, $\lambda$ should also be an eigen value of a square matrix $A^{t}.$
Proof $2$ :- If $\lambda$ is an eigen value of a square matrix $A$ then $1/ \lambda$ is an eigen value of a square matrix $A^{-1}$
$AX=\lambda X$
Pre-multiplying by $A^{-1}$ both sides,
$\Rightarrow X=A^{-1}\lambda X$
$\Rightarrow X=\lambda A^{-1} X$
$\Rightarrow \frac{1}{\lambda}X= A^{-1} X$
So, $1/ \lambda$ should be an eigen value of a square matrix $A^{-1}$