Determinant with Rows Transposed/Proof 1

Theorem

If two rows of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.

Proof

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.

$\blacksquare$

Sources

• 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.9$