Show $\text{det}(A)$ is divisible by $2^{n-1}$ Let $A_{n \times n}$ be a $n$ by $n$ matrix with all entries equal to $\pm 1$. I want to show $\text{det}(A)$ is divisible by $2^{n-1}$. How can I appraoch this?
 A: After one step of Gauß algorithm, the matrix looks like
$$\begin{pmatrix}\pm 1 & \star \\ 0 & B\end{pmatrix},$$
where $B$ is a matrix, whose entries are all in $\{0, \pm 2\}$. Thus you can use Show $\text{det}(A)$ is divisible by $a^n$ .
A: For $n = 1$, the claim is obviously true.

Proceed by induction.

Assume the claim holds for a given positive integer $n$.

Let $A = (a_{ij})$ be an $(n + 1) \times (n+1)$ matrix such $|a_{ij}| = 1$, for all $i,j$.

By successive changes, one entry at a time, change the entries of $A$ so that they all are equal to $+1$. In other words, at each stage, find one entry of $A$, if any, which is equal to $-1$ (it doesn't matter which), and change the value from $-1$ to $+1$, yielding a new matrix $A^\prime$.

Then $|\text{det}(A^\prime) - \text{det}(A)| = |2c_{ij}|$, where $c_{ij}$ is the cofactor of $A$ at row $i$, column $j$.

By the inductive hypothesis, $2^{n-1}\,|\,c_{ij}$, hence $2^n|(\text{det}(A^\prime) - \text{det}(A))$.

Equivalently, $\text{det}(A^\prime) \equiv \text{det}(A) \pmod {2^n}$.

Thus, as the transitions progress, the determinants mod $2^n$ stay the same.

But at the end of all the transitions, the final matrix is the $(n + 1) \times (n+1)$ matrix of all ones, hence (being singular) has deteminant $0$.

It follows that $\text{det}(A) \equiv 0 \pmod {2^n}$, hence $2^n\,|\,\text{det}(A)$, which completes the induction.

