Determinant of Transpose

Theorem
Let $$\mathbf{A} = \left[{a}\right]_{n}$$ be a square matrix of order $n$.

Let $$\det \left({\mathbf{A}}\right)$$ be the determinant of $$\mathbf{A}$$.

Let $$\mathbf{A}^t$$ be the transpose of $$\mathbf{A}$$.

Then $$\det \left({\mathbf{A}}\right) = \det \left({\mathbf{A}^t}\right)$$.

Proof
Let $$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{bmatrix}$$.

Then $$\mathbf{A}^t = \begin{bmatrix} a_{11} & a_{21} & \ldots & a_{n1} \\ a_{12} & a_{22} & \cdots & a_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \cdots & a_{nn} \\ \end{bmatrix}$$.

Let $$b_{rs} = a_{sr}$$ for $$1 \le r, s \le n$$.

We need to show that $$\det \left({\left[{a}\right]_{n}}\right) = \det \left({\left[{b}\right]_{n}}\right)$$.

By the definition of determinant and Permutation of Determinant Indices, we have:

$$ $$ $$

Comment
Thus there is symmetry between rows and columns of a determinant.

So, if we can prove something about a determinant's rows, it follows that this also applies to the determinant's columns, and vice versa.