In how many ways can we fill $n \times n$ matrix with $+1$ or $-1$ such that the product of the entries in each row and each column equals $-1$?
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- $\begingroup$ I have tried upto : In 2*2 matrix No. of ways of filling such arrangement is 2; In 3*3 matrix No. of ways of filling such arrangement is 16; but i have done it by trial and error (i.e, no valid process); so how can i solve this problem $\endgroup$Suresh– Suresh2018-01-22 14:20:21 +00:00Commented Jan 22, 2018 at 14:20
- 2$\begingroup$ Here is a related question. $\endgroup$Rodrigo de Azevedo– Rodrigo de Azevedo2018-04-21 18:44:32 +00:00Commented Apr 21, 2018 at 18:44
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1 Answer
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0 Let the matrix size be $n$.
For each row and each column, there must be an odd number of $(-1)$s. (Why?)
You can arbitrarily assign $1$ and $-1$ for each element of the upper-left $(n-1)\times(n-1)$ cells. The last column will have to balance its previous rows (aka make the row product $-1$), and the last row will have to balance the previous columns (aka make the column product $-1$).
Can you derive a summation expression now?
