how to prove det(B) = − det(A)

Question

Suppose that $A$ is an $n\times n$ matrix and $A −(R_i ↔ R_j)→ B$ is obtained from $A$ by switching rows $R_i$ and $R_j$.

I am not really sure how to prove this. Thanks!

Answer 1

Switching for the rows can be obtained by left matrix multiplication by $S$ that is

$$B=SA $$

with $\det(S)=-1$. Then recall that

$\det(B)=\det(SA )=\det(S)\det(A)=-\det(A)$

Note that $S$ matrix is obtained by $I$ matrix switching the corresponding rows.

Answer 2

You can prove this using the following row operations:

I just show it for $i=1$, $j=2$ to save time:

$$det\begin{pmatrix} r_1 \\ r_2 \\ \vdots \\r_n\end{pmatrix} = det\begin{pmatrix} r_1+r_2 \\ r_2 \\ \vdots \\r_n\end{pmatrix} = - det\begin{pmatrix} r_1+r_2 \\ -r_2 \\ \vdots \\r_n\end{pmatrix} = - det\begin{pmatrix} r_1+r_2 \\ r_1 \\ \vdots \\r_n\end{pmatrix} = - det\begin{pmatrix} r_2 \\ r_1 \\ \vdots \\r_n\end{pmatrix}$$

Answer 3

A determinant can be seen as alternating multilinear form.

So that property is a result of the alternation propery. $$ \DeclareMathOperator{det}{det} \det(\dotsc, a_i, \dotsc, a_j, \dotsc) = -\det(\dotsc, a_j, \dotsc a_i, \dotsc) $$

If you use the Leibniz definition of the determinant $$ \DeclareMathOperator{sgn}{sgn} \det A = \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^n a_{i\pi(i)} $$ it is a property of the sign of the permutations.