Prove that an $n×n$ matrix with entries $+1$ or $-1$ has determinant divisible by $2^{n-1}$

Question

Prove that an $n×n$ matrix with entries $+1$ or $-1$ has determinant divisible by $2^{n-1}$.

Answer 1

Take the first row, and either add or subtract it to each of rows $2$ through $n$ in order to cancel out the first entry in each of those rows to $0$. Doing this does not change the determinant of the matrix. When you have finished, you will have a matrix such that all entries except for the first row are even.

Now if you do cofactor expansion along the first column, you will see that the determinant of the whole matrix is equal to the determinant of the lower right$(n-1)\times (n-1)$ submatrix. Now divide the rows of this submatrix by $2$, one at a time, to see that each time the determinant is divided by $2$ as well. At the end you will be left with an integer matrix, which must therefore have integer determinant, and you will have done a total of $n-1$ divisions, proving that the original determinant was in fact a multiple of $2^{n-1}$.