2

Consider a matrix $A_{n\times n}$ with real entries, denoted by $A=\left(A_{1}A_{2}\dotsm A_{n}\right)$, where $A_{i}$ represents the $i$th column of $A$ for $i=1,\dots,n$. Suppose that the last column, $A_{n}$, is the sum of two vectors $u$ and $v$. Prove that the determinant of $A$ is equal to the sum of determinants obtained by replacing $A_{n}$ with $u$ and $v$ in the matrix:

$$ \det(A) = \det(A_{1},\dots,A_{n-1},u) + \det(A_{1},\dots,A_{n-1},v). $$

I have generated numerous examples of varying sizes using online calculators, and this equality seems to hold true consistently. I managed to prove it for the cases of 2x2 and 3x3, but I genuinely have no idea how to demonstrate the general nxn case. Please help.

Lucas
  • 23
  • The proof depends upon the definition which you use. Maybe you can have a look at this question. – Juliamisto Jan 24 '24 at 03:50
  • @Juliamisto I am considering the determinant of the matrix calculated using Gaussian elimination. I'm looking at the related post, but I understand that it's a more advanced topic than mine. Anyway, thank you. – Lucas Jan 24 '24 at 03:54

1 Answers1

6

Let $(x)_i$ denote the $i$-th component of $x$. Additionally, let $A_{ij}$ represent the cofactor associated with the entry $(i, j)$. Then,

\begin{align*} \det(A) & = (u+v)_1 A_{1n} + (u+v)_2 A_{2n} + \dotsm + (u+v)_n A_{nn} \\ & = (u)_1 A_{1n} + (u)_2 A_{2n} + \dotsm + (u)_n A_{nn} \\ & \quad + (v)_1 A_{1n} + (v)_2 A_{2n} + \dotsm + (v)_n A_{nn} \\ & = \det(A_{1},\dots,A_{n-1},u) + \det(A_{1},\dots,A_{n-1},v). \end{align*}

This is using the method for calculating determinants known as the Rule of Laplace or the Cofactor Method.

https://en.wikipedia.org/wiki/Laplace_expansion

Kevin
  • 735