Suppose $A$ is a general $n \times n$ matrix and $B$ is obtained by interchanging two rows of $A$. Prove that $\det(B) = -\det(A)$

Solution 1:

If $B$ is the matrix obtained by switching rows $i$ and $j$ of $A$, then $B=EA$ where $E$ is the matrix obtained by switching rows $i$ and $j$ of the identity $I$. Clearly $\det E=-1$ and it follows that $$ \det(B)=\det(EA)=\det(E)\det(A)=-\det(A) $$

Solution 2:

$A_{ij}$ is $(n-1)\times (n-1)$ matrix which is a minor from $A$ by deleting $i$-th row and $j$-th column

Recall that $$ {\rm det}\ A =\sum_{j=1}^n (-1)^{i+j} a_{ij}{\rm det}\ A_{ij} $$

If $B$ is from $A$ by exchanging first and second rows (For convenience we consider only this), $$ A_{1j}=B_{2j} $$ so that $${\rm det} \ B = \sum_{j=1}^n (-1)^{2+j} a_{1j} {\rm det}\ B_{2j} = \sum_{j=1}^n (-1)^{2+j} a_{1j} {\rm det}\ A_{1j} =- {\rm det}\ A $$

Solution 3:

Here is a tedious way: \begin{eqnarray} \det B &=& \sum_\sigma \operatorname{sgn} \sigma B_{1 \sigma_1} B_{2 \sigma_2} \cdots B_{n \sigma_n} \\ &=& \sum_\sigma \operatorname{sgn} \sigma A_{2 \sigma_1} A_{1 \sigma_2} \cdots A_{n \sigma_n} \\ &=& \sum_\sigma \operatorname{sgn} \sigma A_{1 \sigma_2} A_{2 \sigma_1} \cdots A_{n \sigma_n} \\ \end{eqnarray} Let $\pi = (1 \ 2)$ be the permutation $\pi_1 = 2, \pi_2 = 1, \pi_k = k$, for $k >2$.

Then $ A_{1 \sigma_2} A_{2 \sigma_1} \cdots A_{n \sigma_n} = A_{1 (\sigma \circ \pi)_1} A_{2 (\sigma \circ \pi)_2} \cdots A_{n (\sigma \circ \pi)_n} $. Furthermore, $\operatorname{sgn} (\sigma \circ \pi) = \operatorname{sgn} \sigma \operatorname{sgn} \pi = - \operatorname{sgn} \sigma$, and so \begin{eqnarray} \sum_\sigma \operatorname{sgn} \sigma A_{1 \sigma_2} A_{2 \sigma_1} \cdots A_{n \sigma_n} &=& - \sum_\sigma \operatorname{sgn} (\sigma \circ \pi) A_{1 (\sigma \circ \pi)_1} A_{2 (\sigma \circ \pi)_2} \cdots A_{n (\sigma \circ \pi)_n}\\ &=& - \sum_{\sigma'} \operatorname{sgn} \sigma' A_{1 \sigma'_1} A_{2 \sigma'_2} \cdots A_{n \sigma'_n}\\ &=& - \det A \end{eqnarray}

Alternatively: Here is a less tedious way. Let $a_1,...,a_n$ be the rows of $A$, then we can consider $\det$ to be a multilinear function of the rows. Then $\det(a_1+a_2,a_1+a_2,a_3,...,a_n) = 0$ since the first two rows are the same. Using the fact that $\det$ is multilinear, we have $$\det(a_1+a_2,a_1+a_2,a_3,...,a_n) = \det(a_1,a_1,a_3,...,a_n) + \det(a_1,a_2,a_3,...,a_n) + \det(a_2,a_1,a_3,...,a_n) + \det(a_2,a_2,a_3,...,a_n)$$ and since the first and third terms are zero, $\det(a_1,a_2,a_3,...,a_n) = - \det(a_2,a_1,a_3,...,a_n)$.