Question

In how many ways can one fill a $m\times n$ table with $\pm 1n$ such that the product of the entries in each row and each column equals-1?

• A. $P\prod _{i=1}^{m-1}{a}_{in}={(-1)}^{m-1}$ hence $(\ast )$ holds if and only if m and n have the same parity.
• B. $P\prod _{i=1}^m{a}_{in}={(-1)}^{m-1}$ hence $(\ast )$ holds if and only if m and n have the same parity.
• C. $P\prod _{i=1}^{m-1}{a}_{in}={\left(1\right)}^{m-1}$ hence $(\ast )$ holds if and only if m and n have the same parity.
• D. $P\prod _{i=1}^{m+1}{a}_{in}={(-1)}^{m-1}$ hence $(\ast )$ holds if and only if m and n have the same parity.

Answer 1

The correct option is A $P\prod _{i=1}^{m-1}{a}_{in}={(-1)}^{m-1}$ hence $(\ast )$ holds if and only if m and n have the same parity.

Denote by $a_{ij}=\pm 1$ in an arbitrary way. This can be done in $2^{(m-1)(n-1)}$ The values for $a_{mj}$ with $1\le j\le n-1$ and for $a_{in}$ with $1\le i\le m-1$ are uniquely determined by the condition that the product of the entries in each row and each column equals $-1$. The value of a $a_{mn}$ is also uniquely determined but it is necessary that ${\prod }_{j=1}^{n-1}amj={\prod }_{i=1}^{m-1}{a}_{in}.$ If we denote $P={\prod }_{i=1}^{m-1}{\prod }_{j=1}^{n-1}{a}_{ij}$ we observe that $P\prod _{j=1}^{n-1}{a}_{mj}={(-1)}^{n-1}$ and $P\prod _{i=1}^{m-1}{a}_{in}={(-1)}^{m-1}$ hence ( ) holds if and only if m and n have the same parity.