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## Permutable line

We say that a line in a matrix $A \in \mathcal{M}_n(\Bbb{R})$ is permutable if permuting its elements does not change the determinant of $A$. Prove that if $A$ has two permutable lines then its determinant is zero.

Proof: First, let’s see what means for a line to be permutable. For the line $k$, which we suppose that is permutable, denote by $a_1,...,a_n$ the elements of the line and $C_1.C_2,..,C_n$ the corresponding cofactors, i.e. $\det(A)=a_1C_1+...+a_nC_n$. This identity will hold for any permutation of $a_i$‘s. Therefore, in the case of transpositions we get $(a_i-a_j)(C_i-C_j)=0$. If all $a_i$ are equal then obviously, the line is permutable. Suppose now that there are $a_i \neq a_j$. This means that $C_i=C_j=C$. Pick another index $l$. Then there exist one of $i,j$ such that $a_l \neq a_i$, for example. Then $C_l=C_i=C$. Doing this for all $l$, we see that all cofactors are equal.
• if the lines each have equal cofactors, then the classical adjoint $A^*$ has two equal lines, which means that $\det(A^*)=\det^n(A)=0$.
• The third case is when we have a line with equal elements $a$ and a line $(b_i)$ with equal cofactors $C$. Then $\det(A)=\sum_i b_iC=C\sum_i b_i$. We can substract $a$ from the $(b_i)$, and the determinant does not change. $\det(A)=\sum_i (b_i-\lambda a)C=C\sum_i (b_i-\lambda a)=0$, for $\lambda = \sum_i b_i /n$.