Multiple of Row Added to Row of Determinant

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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 1

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:

\(\displaystyle \map \det {\mathbf B}\) \(=\) \(\displaystyle \map \det {\mathbf E \mathbf A}\) Determinant of Matrix Product
\(\displaystyle \) \(=\) \(\displaystyle \map \det {\mathbf A}\) as $\map \det {\mathbf E} = 1$

Hence the result.

$\blacksquare$


Proof 2

By Determinant as Sum of Determinants:

$\map \det {\mathbf B} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} + k a_{s1} & a_{r2} + k a_{s2} & \cdots & a_{rn} + k a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} & a_{r2} & \cdots & a_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ k a_{s1} & k a_{s2} & \cdots & k a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}$


Thus by Determinant with Row Multiplied by Constant:

$\map \det {\mathbf B} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} + k a_{s1} & a_{r2} + k a_{s2} & \cdots & a_{rn} + k a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} & a_{r2} & \cdots & a_{rn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} + k \begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix}$


But by Square Matrix with Duplicate Rows has Zero Determinant:

$\begin{vmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{s1} & a_{s2} & \cdots & a_{sn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \\ \end{vmatrix} = 0$


The result follows.

$\blacksquare$


Also see


Sources