Some odd cows

The answer is yes, you can help Leonard. But why, and how ?

Well, we might need to do a little bit of linear algebra here... And the first part, which would have been the hardest if the question was not in the "algebra" section of the website, is to represent the issus with matrix and vectors. Let us give a number from 1 to 23 to the cows, and let us mark as $w_{1},..., w_{23}$ their weights. Let us also mark as W the vector $\begin{pmatrix} w_{1} \\ w_{2} \\ ... \\ w_{23} \end{pmatrix}$ . Our goal is to find W.

The idea is to translate what Albert told us into the equation $M \times W = 0$ where M is a matrix of size 23*23 that represents the data we have. The line i of matrix M will represent the data we gain by setting appart the cow i from the others. We know that we can create 2 groups A and B from the remaining cows of 11 cows each, so that $\sum_{k \, in\, A} w_{k} - \sum_{j \, in\, B} w_{j} = 0$ and $A \cap B = \varnothing$ . However we do not know A and B... We can represent this sum in M by giving to $M_{i,i}$ the value 0 (the cow i does not appear in the sum), to 11 of the $M_{i, k}$ the value +1 (group A from the sum), and to the 11 remaining $M_{i,j}$ the value -1 (group B from the sum). We do not know which cows are in group A and which are in group B, so we will in our chain of reasonning consider all the matrix M that could have been created by this process, and try to find a general result on this set of matrices.

Now, we will introduce a new matrix, called $N_{23}$ . If $J_{23}$ is the matrix of size 23*23 full of 1s, and $I_{23}$ is the identity matrix, we have $N_{23} = J - I$ . We find that $N_{23}$ 's determinant is 22 (even), and that $N_{22}$ 's determinant is -21 (odd).

Now please observe that the parity of a matrix M's determinant does not change if we modify M so that each of its coefficient keeps the same parity (this is due to Gauss's theorem and the definition of the determinant as a sum). Since all of the matrix M in our set of possible representation of the problem can be changed to $N_{23}$ while keeping each coefficient's parity, we know that any matrix in our set has an even determinant. But we already knew that, because it is obvious that for any M in our set, the vector $V = \begin{pmatrix} 1 \\ 1 \\ ... \\ 1 \end{pmatrix}$ is a solution to the equation $M \times X = 0$ . Therefore, M is not inversible, and its determinant is 0 (even).

What we learned from transforming all the Ms in Ns, is that the matrix of size 22*22 obtained by getting rid of the last line and last column of any M in our set has an odd determinant, and is therefore inversible. This means that the rank of any M in our set is at least 22 (a submatrix is inversible), but at most 23 (size of M), and is not 23 (M is not inversible). It has to be 22 !

This is great, because by the rank-nullity theorem, we know that any M in our set has a null space of dimension 1. We already noticed that V was always a non-null vector of this null space, we have therefore proven that the null space of any matrix in our set is the span of V. Hence, W is in the span of V !

What does it mean for our question ? Well, we know that any set of cows that verifies Albert's property, is a set in which all the cows have the very same weight. And we know that one of the cows weights 400 kgs. Hence, all cows weight 400 kgs ! I bet Leonard will be happy, and impressed by your reasonning !