Determinant with Rows Transposed

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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:

\(\displaystyle \map \det {\mathbf B}\) \(=\) \(\displaystyle \map \det {\mathbf E \mathbf A}\) Determinant of Matrix Product
\(\displaystyle \) \(=\) \(\displaystyle -\map \det {\mathbf A}\) 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