Proof

We first prove when \(n=3\) whose matrix of two rows are adjacent.

If we expand the determinant of LHS across the 2nd row, we have

$$\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right|=(-1)^{2+1}a_{21}\left|\begin{array}{cc}a_{12}& a_{13}\\ a_{32}& a_{33}\end{array}\right|+(-1)^{2+2}a_{22}\left|\begin{array}{cc}a_{11}& a_{13}\\ a_{31}& a_{33}\end{array}\right|+(-1)^{2+3}a_{23}\left|\begin{array}{cc}a_{11}& a_{12}\\ a_{31}& a_{32}\end{array}\right|$$

If we expand the determinant of RHS across the 3rd row, we have

$$-\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{31}& a_{32}&a_{33}\\a_{21}& a_{22}&a_{23}\end{array}\right|=-\left((-1)^{3+1}a_{21}\left|\begin{array}{cc}a_{12}& a_{13}\\ a_{32}& a_{33}\end{array}\right|+(-1)^{3+2}a_{22}\left|\begin{array}{cc}a_{11}& a_{13}\\ a_{31}& a_{33}\end{array}\right|+(-1)^{3+3}a_{23}\left|\begin{array}{cc}a_{11}& a_{12}\\ a_{31}& a_{32}\end{array}\right|\right)$$

Therefore, two expressions are equal. ( General case of n, we can prove by expanding LHS across \(i\)-th row and RHS across \((i+1)\)-row.)

Thus if adjacent two rows of a matrix are interchanged, the sign of its determinant is changed:

$$\left|\begin{array}{cccc}a_{11}& a_{12}&\cdots &a_{1n}\\\vdots &\vdots &\vdots& \vdots\\a_{i1}& a_{i2}&\cdots &a_{in}\\a_{i+1,1}& a_{i+1,2}&\cdots &a_{i+1,n}\\\vdots &\vdots &\vdots& \vdots\\a_{n1}& a_{n2}&\cdots&a_{nn}\end{array}\right|=-\left|\begin{array}{cccc}a_{11}& a_{12}&\cdots &a_{1n}\\\vdots &\vdots &\vdots& \vdots\\a_{i+1,1}& a_{i+1,2}&\cdots &a_{i+1,n}\\a_{i1}& a_{i2}&\cdots &a_{in}\\\vdots &\vdots &\vdots& \vdots\\a_{n1}& a_{n2}&\cdots&a_{nn}\end{array}\right|$$

Next, we see the case when \(i\)-th row and \(j\)-th row are interchanged.

In such a case, we need to interchange the rows \(2(j-i)-1\) times ( \(i<j\) ).

Therefore, the sign of the determinant is changed by

$$(-1)^{2(j-i)-1}=(-1)^{-1}=-1$$

Thus, we can conclude that if any two rows of a matrix are interchanged, the sign of its determinant is changed :

$$\left|\begin{array}{cccc}a_{11}& a_{12}&\cdots &a_{1n}\\\vdots &\vdots &\vdots& \vdots\\a_{i1}& a_{i2}&\cdots &a_{in}\\\vdots &\vdots &\vdots& \vdots\\a_{j1}& a_{j2}&\cdots &a_{jn}\\\vdots &\vdots &\vdots& \vdots\\a_{n1}& a_{n2}&\cdots&a_{nn}\end{array}\right|=-\left|\begin{array}{cccc}a_{11}& a_{12}&\cdots &a_{1n}\\\vdots &\vdots &\vdots& \vdots\\a_{j1}& a_{j2}&\cdots &a_{jn}\\\vdots &\vdots &\vdots& \vdots\\a_{i1}& a_{i2}&\cdots &a_{in}\\\vdots &\vdots &\vdots& \vdots\\a_{n1}& a_{n2}&\cdots&a_{nn}\end{array}\right|$$