Proof

Suppose that A is an \(n\times n\) matrix as follows:

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

Let us denote the \((i,j)\)-entry in A by \([A]_{ij}\).

Then we have

$$\begin{eqnarray*}&&[A]_{ij}=a_{ij}\\&&[A^{t}]_{ij}=[A]_{ji}=a_{ji}=:a’_{ij}\end{eqnarray*}$$

Then if we expand the determinant of A across \(i\) th row, we have

$$|A|=\displaystyle\sum_{j=1}^{n}(-1)^{i+j}a_{ij}|A_{ij}|$$

where \(A_{ij}\) represents the submatrix formed by delating the \(i\) th row and \(j\) th column of A.

Similarly, if we expand the determinant of \(A^{t}\) across \(i\) th column, we have

$$\begin{eqnarray*}|A^{t}|&=&\displaystyle\sum_{k=1}^{n}(-1)^{k+i}a’_{ki}|A^{t}_{ki}|\\&=&\displaystyle\sum_{k=1}^{n}(-1)^{i+k}a_{ik}|A_{ik}|\ \ \ \text{( ∵ \(a’_{ki}=a_{ik}\) and \(|A^{t}_{ki}|=|A_{ik}|\) )}\end{eqnarray*}$$

where \(A^{t}_{ki}\) is the submatrix formed by delating the \(k\) th row and \(i\) th column of \(A^{t}\), whose determinant is equivalent to the determinant of \(A_{ik}\).

These equations are equal. Therefore, we obtain

$$|A|=|A^{t}| $$