Theorem 4.8
For any \(A \in M_{n \times n}(\mathbf{F})\), \(\det(A^t) = \det(A)\).


Proof:

If \(A\) is not invertible, then \(\text{rank}(A) < n\). We know that \(\text{rank}(A) = \text{rank}(A^t)\) by theorem 3.6 (corollary 2). Therefore, \(A^t\) is not invertible and \(\det(A) = \det(A^t) = 0\).

Suppose now that \(A\) is invertible, then we can write \(A\) as a product of elementary matrices, \(A = E_k,E_{k-1}...,E_1\). Furthermore, we’ve proved that \(\det(E) = \det(E^t)\) for any elementary matrix. From this, notice that

$$ \begin{align*} \det(A^t) &= \det((E_k,E_{k-1}...,E_1)^t) \\ &= \det(E_1^tE_2^t...E_k^t) \\ &= \det(E_1^t)\det(E_2^t)...\det(E_k^t) \\ &= \det(E_1)\det(E_2)...\det(E_k) \\ &= \det(E_k)\det(E_{k-1})...\det(E_1) \\ &= \det(E_k,E_{k-1}...,E_1) \\ &= \det(A). \ \blacksquare \end{align*} $$



References: