Step 1: Restate What Orthogonal Means.
A matrix with $A^{T} = A^{-1}$ preserves the length of every vector it acts on: for any vector $x$, $\|Ax\|^2 = (Ax)^{T}(Ax) = x^{T}A^{T}Ax = x^{T}A^{-1}Ax$. Since $A^T=A^{-1}$, this simplifies to $x^Tx = \|x\|^2$. So $A$ is a length-preserving (orthogonal) transformation, such as a rotation or reflection.
Step 2: Use the Product-of-Eigenvalues Rule.
The determinant of any square matrix equals the product of its eigenvalues, $\det(A) = \lambda_1 \lambda_2 \cdots \lambda_n$. For an orthogonal matrix, every eigenvalue (allowing complex ones) has magnitude 1, because the transformation does not stretch or shrink vectors. Complex eigenvalues of a real matrix always occur in conjugate pairs $\lambda, \bar\lambda$, and each such pair multiplies to $|\lambda|^2 = 1$.
Step 3: Combine the Eigenvalues.
So the full eigenvalue product is a chain of factors that are each $+1$ (real eigenvalues equal to $+1$), $-1$ (real eigenvalues equal to $-1$), or $1$ (conjugate pairs). Multiplying any combination of $+1$'s and $-1$'s together always gives either $+1$ or $-1$, never any other real number. Since $\det(A)$ is real (A is a real matrix), it must land on exactly one of these two values.
Step 4: Rule Out n and 0.
A determinant of $n$ would need the eigenvalue magnitudes to depend on the matrix size in a way that breaks the length-preserving property proved in Step 1, so it cannot happen. A determinant of $0$ would mean $A$ is not invertible, but the problem states $A^{-1}$ exists (it equals $A^T$), so $0$ is impossible.
Final Answer:
$\det(A)$ can only be $+1$ or $-1$.
\[ \boxed{+1 \text{ and } -1 \text{ i.e. options (A) and (B)}} \]