A $4\times4$ determinant with entries $\pm1$ is divisible by 8 Let B be a 4 by 4 matrix all of whose entries are -1 or 1
show det(B) is divisible by 8
Anyone can guide me for this?
Thanks!
 A: Multiplying rows and columns of $B$ by $-1$ as necessary, which at most changes the sign of $\det(B)$, we can assume
$$B=\pmatrix{1&1&1&1\\-1&\pm1&\pm1&\pm1\\-1&\pm1&\pm1&\pm1\\-1&\pm1&\pm1&\pm1}$$
Adding the first row to each of the others gives
$$\det(B)=\det\pmatrix{1&1&1&1\\0&2a&2b&2c\\0&2d&2e&2f\\0&2g&2h&2i}=8\det\pmatrix{a&b&c\\d&e&f\\g&h&i}$$
where $a,b,c,d,e,f,g,h,i$ are all $0's$ or $1$'s, but in particular the last determinant is an integer.
A: Fix the 1st (may be any) row of the matrix. If we have any row whose elements are exactly having the sign opposite to the 1st row and since determinant remains unaltered if elements of one row are added to the other so we get one row with zero entries and hence determinant=0 divisible by 8. If not so convince yourself by checking that if the entries of the 2nd 3rd and 4th row are added to the 1st row then the entries in the 2nd,3rd,4th row will become -2, 2 or 0 from where we can take 2 (or -2) common from each row (2nd,3rd,4th) and have the determinant of the matrix divisible by 8
A: This is just a better formulation of learnmore's answer.
Since the matrix has integer coefficients, its determinant is integer, so divisibility makes sense.
Row reduce the matrix, using the entry at $(1,1)$ as pivot. Then, after eliminating under it, we have the matrix in the form
$$
\begin{bmatrix}
\pm1 & \pm1 & \pm1 & \pm1 \\
0 & x_{11} & x_{12} & x_{13} \\
0 & x_{21} & x_{22} & x_{23} \\
0 & x_{31} & x_{32} & x_{33}
\end{bmatrix}
$$
where the entries $x_{ij}$ are $2$, $0$ or $-2$. The necessary operations don't change the determinant of the matrix and only require integers (summing the first row, possibly multiplied by $-1$, with one of the others). The matrix
$$
\begin{bmatrix}
x_{11} & x_{12} & x_{13} \\
x_{21} & x_{22} & x_{23} \\
x_{31} & x_{32} & x_{33}
\end{bmatrix}
$$
has determinant divisible by $8$, because we can collect $2$ from each row (or column) still getting a matrix with integer coefficients.
Note that the same argument shows that an $n\times n$ matrix with $\pm1$ coefficients has determinant divisible by $2^{n-1}$ (for $n>1$, of course).
