Let us consider a square matrix of order $n$ with entries $1$ or $-1$ having product of each row and column equal to $-1$. Then total number of such matrices is...
Number of matrices with product of entries -1
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combinatorics
matrices
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0..is something that we have to calculate. Try calculating it for small values of $n$. Spot a pattern. Try induction. – 2017-01-31
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0Suppose you fix the value of the upper-left $(n-1)\times(n-1)$ submatrix. How many ways are there to fill in the last row and column so that the full matrix has the desired property? – 2017-01-31
1 Answers
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see The matrix should have odd number of $(-1) 's$ in each row and you can do this for the $(n-1)$ rows .But the last row you have no choice because that is governed by the preceeding product. So number of matrices should be
$\sum_{k=0}^{ \lfloor \frac{n-1}{2}\rfloor} {n\choose 2k+1}.(n-1)$