Determinant with Rows Transposed
Theorem
If two rows of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.
Proof 1
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $1 \le r < s \le n$.
Let $e$ be the elementary row operation that exchanging rows $r$ and $s$.
Let $\mathbf B = \map e {\mathbf A}$.
Let $\mathbf E$ be the elementary row matrix corresponding to $e$.
From Elementary Row Operations as Matrix Multiplications:
- $\mathbf B = \mathbf E \mathbf A$
From Determinant of Elementary Row Matrix: Exchange Rows:
- $\map \det {\mathbf E} = -1$
Then:
\(\ds \map \det {\mathbf B}\) | \(=\) | \(\ds \map \det {\mathbf E \mathbf A}\) | Determinant of Matrix Product | |||||||||||
\(\ds \) | \(=\) | \(\ds -\map \det {\mathbf A}\) | as $\map \det {\mathbf E} = -1$ |
Hence the result.
$\blacksquare$
Proof 2
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $1 \le r < s \le n$.
Let $\rho$ be the permutation on $\N^*_n$ which transposes $r$ and $s$.
From Parity of K-Cycle, $\map \sgn \rho = -1$.
Let $\mathbf A' = \sqbrk {a'}_n$ be $\mathbf A$ with rows $r$ and $s$ transposed.
By the definition of a determinant:
- $\displaystyle \map \det {\mathbf A'} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a'_{k \map \lambda k} }$
By Permutation of Determinant Indices:
- $\displaystyle \map \det {\mathbf A'} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$
We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
- $\displaystyle \map \det {\mathbf A'} = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} } = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$
Hence the result.
$\blacksquare$
Also see
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): determinant (1)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): determinant (1)
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): determinant (ii)