2

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

By general $n \times n$ matrix, I mean $$\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{matrix} \right]$$ I'm thinking of using the cofactor expansion of the 2nd row of the first matrix, yielding $$ \det(A) = c_{21} a_{21} + c_{22} a_{22} + \cdots+ c_{2n} a_{2n} $$ which will be positive or negative depending on $n$ and the values of $a_{21}, \ldots,a_{2n}$, and using the same on the first row in matrix $B$ to get $c_{11} a_{21}$ etc. and then use the fact that the power of the $c$ subscripts will alter $\det(B)$ to be the opposite sign, but I'm not sure how to express all of this mathematically in a proof directly. Help please?

Martin Sleziak
  • 53,687
J L
  • 1,405

4 Answers4

9

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 $$ \det(B)=\det(EA)=\det(E)\det(A)=-\det(A) $$

Brian Fitzpatrick
  • 26,559
4

$A_{ij}$ 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 $${\rm det} \ B = \sum_{j=1}^n (-1)^{2+j} a_{1j} {\rm det}\ B_{2j} = \sum_{j=1}^n (-1)^{2+j} a_{1j} {\rm det}\ A_{1j} =- {\rm det}\ A $$

HK Lee
  • 19,964
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, $\det(a_1,a_2,a_3,...,a_n) = - \det(a_2,a_1,a_3,...,a_n)$.

copper.hat
  • 172,524
0

if the matrix is singular there is nothing to prove. otherwise use linear operations to reduce the entries in the first two columns to zeroes, except for the first two entries in each. the determinant is unaltered by these operations. now $$ |A| = |A_2||A_{n-2}| $$ interchanging the first two rows gives $$ |A^*|=|A_2^*||A_{n-2}| $$ and evidently $|A_2^*|=-|A_2|$

David Holden
  • 18,040