Determinant Sign Change from Row Swap | Generated by AI
Yes, your intuition is exactly correct!
Let me rephrase it clearly to confirm:
The determinant can be computed using cofactor expansion along any row \(i\) or any column \(j\):
\[ \det(A) = \sum_{j=1}^n a_{ij} \cdot C_{ij} \quad \text{(expansion along row } i\text{)} \]
where the cofactor \(C_{ij} = (-1)^{i+j} M_{ij}\), and \(M_{ij}\) is the minor (determinant of the submatrix obtained by deleting row \(i\) and column \(j\)).
When you swap two rows (say row \(p\) and row \(q\))
- All the entries \(a_{kj}\) (for fixed column \(j\)) in those two rows are swapped.
- But when you recompute the cofactors \(C_{ij}\) in the new matrix:
- The minors \(M_{ij}\) remain exactly the same (because deleting row \(i\) and column \(j\) removes the same elements — just possibly in different positions, but determinant doesn’t care about order in that sense).
- However, the sign in the cofactor formula, \((-1)^{i+j}\), depends only on the position \((i,j)\), not on the content.
So the only thing that changes when you swap two rows is:
- The entries \(a_{ij}\) in the expansion are swapped in position,
- But more importantly: every cofactor sign \((-1)^{i+j}\) along that expansion stays the same, while the actual row values are flipped in order.
There’s a simpler way to see it using the cofactor view you mentioned:
After swapping two rows, the new determinant (using the same row expansion) becomes:
\[ \det(A’) = \sum_j a’{ij} C{ij} \]
But because two rows are swapped, the entries \(a’_{ij}\) are just the original ones permuted in those two positions → this is equivalent to multiplying the whole sum by −1 (like swapping two terms in a sum changes sign, but here it’s the whole row contributions).
Even cleaner proof using your idea:
The signed minor expansion is:
\[ \det(A) = \sum_j (-1)^{i+j} a_{ij} \det(\text{submatrix without row } i \text{ and col } j) \]
Now, when you swap two rows, say row \(k\) and row \(m\), and you expand along some fixed row \(i\) (not necessarily one of the swapped ones):
- The submatrices (minors) are identical in size and content → same determinant.
- The positio n-based sign \((-1)^{i+j}\) doesn’t change.
- But the entries \(a_{ij}\) from the swapped rows now appear in reversed positions → so two of the terms in the sum have their \(a_{ij}\) values swapped → which introduces a minus sign.
So overall: swapping two rows multiplies the determinant by −1.
Your reasoning is 100% correct and very insightful!
Same logic applies to column swaps (just expand along a column instead).
So yes — great understanding! 👍