# How many matrices in which product of numbers in each row and column is 1?

We have a matrix $$S$$. Every $$S_{ij}$$ is equal to $$1$$ or $$-1$$. How many matrices we can have so that product of all entries in every row and column is $$1$$?

This means number of $$-1$$ must be even in all rows and columns. And size of $$S$$ is $$m\times n$$.

My idea was that we can divide a matrix of $$(m-1)\times (n-1)$$, then count all the possible ways this matrix can have $$1$$ and $$-1$$ which is $$2^{(m-1)(n-1)}$$. Then add other part of matrix and set those values in a way that number of $$-1$$ become even. But this way doesn't work.

Sorry if I didn't explain my way very clearly, it's because English is not my native language.

• Two hints : How many $-1$ required in any row or column? You have almost solved the problem. Oct 7, 2020 at 18:30

Consider an example : $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline 1 & -1 & -1 & -1 & 1 & 1 & -1 & \color{blue}{1}\\ \hline 1 & -1 & 1 & -1 & 1 & -1 & 1 & \color{blue}{-1}\\ \hline -1 & 1 & 1 & 1 & 1 & -1 & -1 & \color{blue}{-1}\\ \hline -1 & 1 & -1 & -1 & 1 & -1 & 1 & \color{blue}{1}\\ \hline 1 & 1 & -1 & -1 & -1 & 1 & 1 & \color{blue}{-1}\\ \hline \color{blue}{1} & \color{blue}{1} & \color{blue}{-1} & \color{blue}{1} & \color{blue}{-1} & \color{blue}{-1} & \color{blue}{1} & \color{red}{-1}\\ \hline \end{array}$$

The table is a valid $$m\times n$$ matrix. Black entries denote $$(m-1)\times (n-1)$$ matrix.

Key observation :

• Last entry (in any row/column) = Product of all other entries (in respective row/column)

So all the last entries (colored) are uniquely determined, once entries of $$(m-1)\times (n-1)$$ matrix are arbitrarily chosen. Hence answer is $$2^{(m-1)(n-1)}$$

Note the peculiarity about corner entry. Product of last column blue entries = Product of all black entries = Product of last row blue entries = Red entry

• but how can we be sure the last row and column are valid? maybe the last row has even -1s and last column has odd, what will be the last number? Oct 8, 2020 at 18:27