How would you prove that exchanging two rows of an $n \times n$ matrix $A$ multiplies its determinant by $-1$?

Question

Do you need other results to prove this? Could you use proof by induction?

I don't know where to begin regarding how to prove this.

Answer 1

The determinant can be defined as (Using Einstein's summation notation):

$\det(\mathbf{A}) = \varepsilon_{i_1\dots i_n} a_{1i_1} \dots a_{ki_k} \dots a_{mi_m} \dots a_{ni_n}$

Therefore exchanging the files k and m

$\det(\mathbf{A'}) = \varepsilon_{i_1\dots i_k\dots i_m\dots i_n} a_{1i_1} \dots a_{mi_k} \dots a_{ki_m} \dots a_{ni_n}$

Using the antisymmetry of the levi-civita tensor and renaming the dummy index you get:

$\det(\mathbf{A'}) = -\varepsilon_{i_1\dots i_m\dots i_k\dots i_n} a_{1i_1} \dots a_{mi_k} \dots a_{ki_m} \dots a_{ni_n}=-\varepsilon_{i_1\dots i_k\dots i_m\dots i_n} a_{1i_1} \dots a_{mi_m} \dots a_{ki_k} \dots a_{ni_n}=-\det(\mathbf{A})$

Answer 2

Let $A'$ be $A$ with exchanged $i$-th and $j$-th column.

Let $I'$ be the $n * n$ identity matrix with $i$-th and $j$-th rows exchanged. Using both Leibniz or Laplace determinant definition it's easy to prove that $\det(I') = -1$.

Because exchanging $i$-th and $j$-th column of matrix $A$ is basically multiplying matrix $A$ by $I'$ so $\det(A') = \det(A \cdot I')$, we get $$\det(A') = \det(A \cdot I') = \det(A) \cdot \det(I') = -det(A)$$

We proved that exchanging two columns multiplies determinant by $-1$. Because $\det(A) = \det(A^T)$, so for exchanging rows it's the same.