# Does swapping columns of a matrix cause the rows of the inverse matrix to be swapped?

This question came up while I was performing some computation on a few matrices on an unrelated problem in computer science.

Let $$A$$ be an invertible matrix with columns $$A_1, \dots A_n$$. Let $$B$$ be its inverse, with rows $$B_1, \dots, B_n$$. Now construct a new matrix $$\hat{A}$$ by swapping two columns $$A_i$$ and $$A_j$$. Let $$\hat{B}$$ be the inverse of $$\hat{A}$$.

In my specific computations, I noticed that $$\hat{B}$$ was nothing more than $$B$$ with the rows $$B_i$$ and $$B_j$$ swapped. Is this always true? I doubt it was a coincidence because the numbers were quite random.

Note: I was working with real numbers but I'd be interested to know if the field makes any difference.

Yes, this is always true. Note that swapping columns $$i$$ and $$j$$ is equivalent to multiplying on the right side by the elementary matrix $$T_{ij}$$ which is defined by swapping rows $$i$$ and $$j$$ of the identity matrix. You can check that this matrix is the inverse of itself. Also, multiplying by $$T_{ij}$$ on the left side is equivalent to swapping rows $$i$$ and $$j$$.
So if we call your matrix $$A$$ then $$AT_{ij}$$ is the matrix that you get that swapping columns $$i$$ and $$j$$. Then its inverse is $$T_{ij}A^{-1}$$ which is the matrix that you get when you swap rows $$i$$ and $$j$$ in $$A^{-1}$$.
Another way to think of it: you are just re-numbering the basis for one of the vector spaces: The domain space for $$A$$, which is the range space fo $$A^{-1}$$.
Your observation is spot on. You can swap columns of a matrix by right-multiplying by a permutation matrix $$P$$ that is the identity matrix with the corresponding columns swapped: $$\hat A = AP$$. We then have $$\hat A^{-1} = (AP)^{-1} = P^{-1}A^{-1}.$$ $$P$$ is its own inverse, and left-multiplying by $$P$$ swaps the rows of $$A^{-1}$$ that are swapped in $$P$$.
• $P$ is its own inverse in this specific case because we are swapping columns. In general the inverse of a permutation matrix $P$ is its transpose. – Luca Citi Sep 30 '18 at 19:16