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I'm trying to arrive at the computational definition of the determinant of a square matrix $M$ by first defining it as the signed volume of applying the corresponding linear transformation to the unit cube.

In doing so it seems essential to understand the effect each elementary row operation has on the determinant.

What geometrical argument could then be made to show that adding a scalar multiple of row $i$ to row $j$ does not change the determinant?

I would appreciate any help.

Sam
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    Draw a parallelogram $0, v, w, v + w$ in $\mathbb{R}^2$; the (signed) area of this parallelogram is the determinant of the matrix formed by $v, w$. What happens to the area of the parallelogram if you add a multiple of $v$ to $w$? – Qiaochu Yuan Sep 19 '19 at 06:26
  • Related: https://math.stackexchange.com/questions/427528/why-determinant-is-volume-of-parallelepiped-in-any-dimensions, https://math.stackexchange.com/questions/29128/why-determinant-of-a-2-by-2-matrix-is-the-area-of-a-parallelogram – Hans Lundmark Sep 19 '19 at 06:31

1 Answers1

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Since the determinant is a multilinear map on its rows, we have: $$M = [R_1,...,R_n]^T$$ $$\det [R_1,...,R_n+sR_1]^T = \det [R_1,...,R_n]^T + s\det [R_1,...,R_1]^T = \det(M) + 0 = \det(M)$$ The second $\det$ annihilate because it has a repeated row.

For $i,j$ it's the same.

Ali Ashja'
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