Multiple of Row Added to Row of Determinant/Proof 1
Theorem
Let $\mathbf A = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r 1} & a_{r 2} & \cdots & a_{r n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$ be a square matrix of order $n$.
Let $\map \det {\mathbf A}$ denote the determinant of $\mathbf A$.
Let $\mathbf B = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r 1} + k a_{s 1} & a_{r 2} + k a_{s 2} & \cdots & a_{r n} + k a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s 1} & a_{s 2} & \cdots & a_{s n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end{bmatrix}$.
Then $\map \det {\mathbf B} = \map \det {\mathbf A}$.
That is, the value of a determinant remains unchanged if a constant multiple of any row is added to any other row.
Proof
Let $e$ be the elementary row operation that adds $k$ times row $r$ to row $s$.
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: Scale Row and Add:
- $\map \det {\mathbf E} = 1$
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 $1.11$