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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.

Romanian National Olympiad, 11th grade

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.

Now, if we have two lines which are permutable, there are three cases:

  • the two lines contain equal elements, which mean that the determinant is obviously zero.
  • 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.
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