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I have written this proof stating that if two rows or columns of a matrix are the same, then the determinant of the matrix is equal to 0. Is it correct?

Let us say we have an n x n matrix A, shown below:

n x n matrix A

For some $i,n \in \mathbb{N}$.

If we say that $r_{i} = [a_{i1}, a_{i2}, a_{i3} ... a_{in}]$, a row vector, then we can rewrite the matrix A as:

n x n matrix A 2

If we then also create the swapped $S_{ij}$ matrix, i.e. swap rows $i$ and $j$ around, we have the matrix:

Matrix S

We know that if we swap two rows of a determinant, in this case rows $i$ and $j$, then the determinant will simply be the negative of the original determinant.We can say that the det(A) = -det(S). But, if these two rows are identical, then det(S) = det(A), so this means that det(A) = -det(A), so A must be equal to $0$.

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  • $\begingroup$ Yes, your solution is correct. $\endgroup$
    – Ben Grossmann
    Jun 26, 2020 at 8:12
  • $\begingroup$ It's easier to use that $\det$ is an alternating multilinear form, i.e., $$\det(a_1,\dots,a_i,\dots,a_j\dots,a_n)=-\det(a_1,\dots,a_j,\dots,a_i\dots,a_n).$$ $\endgroup$
    – Michael Hoppe
    Jun 26, 2020 at 9:18

4 Answers 4

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That seems to be correct. Another way to think about this is to consider what happens when there are two identical columns/rows. Then, the number of linearly independent columns/rows is less than $n$ for a given $n \times n$ matrix. Hence, the matrix does not have maximal rank so the determinant has to be 0.

Of course, I'm working off of the idea that you've defined the determinant as the unique map $det: M(n \times n, \mathbb{F}) \to \mathbb{F}$ such that:

  1. $det$ is linear in each row.

  2. If $rank(A) < n$, then $det(A) = 0$.

  3. $det(E_n) = 1$.

If you're not working off of these assumptions, that's fine. What you have written is fine too.

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In essence, yes, your proof is very correct, and thorough.

For further clarification you can see this video https://www.khanacademy.org/math/linear-algebra/matrix-transformations/determinant-depth/v/linear-algebra-duplicate-row-determinant which gives a very thorough breakdown of your proof.

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Yes, this is indeed a valid proof which you can see on occasion in published works.

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If you have two (or more) rows which are linearly related, you can resolve some rows to zero. (Check out gaussian elimination to see how row reduction works.) By the way this proof is valid indeed.

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