Determinant with Column Multiplied by Constant
Theorem
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 $\mathbf B$ be the matrix resulting from one column of $\mathbf A$ having been multiplied by a constant $c$.
Then:
- $\map \det {\mathbf B} = c \map \det {\mathbf A}$
That is, multiplying one column of a square matrix by a constant multiplies its determinant by that constant.
Proof
Let:
- $\mathbf A = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 r} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 r} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\ \end{bmatrix}$
- $\mathbf B = \begin{bmatrix} b_{1 1} & b_{1 2} & \cdots & b_{1 r} & \cdots & b_{1 n} \\ b_{2 1} & b_{2 2} & \cdots & b_{2 r} & \cdots & b_{1 n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n r} & \cdots & b_{n n} \\ \end{bmatrix} = \begin{bmatrix} a_{1 1} & a_{1 2} & \cdots & c a_{1 r} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & c a_{2 r} & \cdots & a_{1 n} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\ \end{bmatrix}$
We have that:
- $\mathbf A^\intercal = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r 1} & a_{r 2} & \cdots & a_{r n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$
where $\mathbf A^\intercal$ denotes the transpose of $\mathbf A$.
Similarly, we have that:
- $\mathbf B^\intercal = \begin{bmatrix} a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\ a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \\ \end {bmatrix}$
From Determinant with Row Multiplied by Constant:
- $\map \det {\mathbf B^\intercal} = c \map \det {\mathbf A^\intercal}$
From from Determinant of Transpose:
- $\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
- $\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$
and the result follows.
$\blacksquare$
Also see
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.10 \ (3)$
- 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: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace: Proposition $1.13$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): determinant (2)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): determinant (2)