Determinant Property for $A^T = A^{-1}$

We are given a square matrix $A$ such that $A^T = A^{-1}$. This condition defines an orthogonal matrix.

To find the possible values for the determinant of $A$, we start with the given equation:

$ A^T = A^{-1} $

Taking the determinant of both sides, we get:

$ \det(A^T) = \det(A^{-1}) $

Applying Determinant Properties

We use two fundamental properties of determinants:

• The determinant of a matrix transpose is equal to the determinant of the original matrix: $\det(A^T) = \det(A)$.
• The determinant of the inverse of a matrix is the reciprocal of the determinant of the original matrix: $\det(A^{-1}) = \frac{1}{\det(A)}$.

Substituting these properties into our equation $\det(A^T) = \det(A^{-1})$, we obtain:

$ \det(A) = \frac{1}{\det(A)} $

Solving for Determinant Value

To solve for $\det(A)$, we multiply both sides by $\det(A)$ (note that $\det(A)$ cannot be 0, otherwise $A^{-1}$ would not exist):

$ (\det(A))^2 = 1 $

Taking the square root of both sides yields the possible values for the determinant:

$ \det(A) = \pm 1 $

Therefore, the determinant of a matrix $A$ satisfying $A^T = A^{-1}$ can only be +1 or -1.

Conclusion on Options

• Option 1 (+1): Possible.
• Option 2 (-1): Possible.
• Option 3 ($n$): Not necessarily true. The dimension $n$ does not directly determine the determinant in this case. For example, a 2x2 rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ has $A^T = A^{-1}$ and its determinant is $\cos^2\theta + \sin^2\theta = 1$. Here $n=2$, but the determinant is 1, not 2.
• Option 4 (0): Not possible, as $\det(A) = 0$ implies the matrix is not invertible, contradicting $A^{-1}$ existing.

The possible values for the determinant are +1 and -1.