Multiple of Row Added to Row of Determinant
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:
\(\ds \map \det {\mathbf B}\) | \(=\) | \(\ds \map \det {\mathbf E \mathbf A}\) | Determinant of Matrix Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \det {\mathbf A}\) | as $\map \det {\mathbf E} = 1$ |
Hence the result.
$\blacksquare$
Proof 2
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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercises -- Second Set
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): determinant (3)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): determinant (3)
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): determinant (iii)