Showing that Matrix $A \in M_{50}(\mathbb{R})$ is invertible I have matrix $A \in M_{50}(\mathbb{R})$. This matrix has $A_{i,i} = 0$ for all $ 1 \le i \le 50$ and $A_{i,j} \in \{\pm1\}$ for all distinct $1 \le i, j \le 50$. I must show that $A$ is invertible. 
In order to do so, I probably would want to show that $det(A)$ is non-zero. However, this is a 50 x 50 matrix, and surely, there must be another way to show that $det(A)$ is non-zero besides brute force computation. Someone told me to consider the image of $A$ in $M_{50}(F_{2})$, but I am not sure how to apply this into the proof. Can anybody clarify some things? Thank you. 
 A: Its determinant is an odd number, so is nonzero.
To see this, let $n$ be an even number (here $n$ is $50$) and let
$M$ be a $n\times n$ matrix with zeros on the diagonal and $\pm 1$s
elsewhere. Then $M\equiv J-I\pmod 2$ where $J$ is the all-one matrix. Therefore
$\det A\equiv\det(J-I)=\det(I-J)\pmod 2$. The characteristic polynomial of $J$ is
$$\det(tI-J)=t^{n-1}(t-nI)$$
since it has rank $1$ and the nonzero eigenvalue $n$. Therefore $\det(I-J)=1-n$,
which is an odd number.
A: Here is a proof along the lines of the hint you were given: considering $A \in M_{50}(\mathbb{Z})$, the image of $A$ in $M_{50}(\mathbb{Z} / 2\mathbb{Z})$ is the matrix with $1 + 2\mathbb{Z}$ in each nondiagonal position and $0 + 2\mathbb{Z}$ in each diagonal position.  Let us call this image $B \in M_{50}(\mathbb{F}_2)$.
Now, we will show that $B$ is an invertible matrix.  To see this, suppose we have $x \in \mathbb{F}_2^{50}$ with $Bx = 0$.  Then from the first row, we have $x_2 + x_3 + \cdots + x_{50} = 0$, so $x_1 = x_1 + x_2 + x_3 + \cdots + x_{50}$.  Similarly, the $i$th row gives $x_i = x_1 + x_2 + x_3 + \cdots + x_{50}$ for each $i$.  Therefore, $x_1 = x_2 = \cdots = x_{50} = x_1 + x_2 + \cdots + x_{50}$.  But $x_1 = x_2 = \cdots = x_{50} = 1 + 2\mathbb{Z}$ will contradict the last equation; therefore, the only remaining possibility is that $x_1 = x_2 = \cdots = x_{50} = 0 + 2\mathbb{Z}$.  We have thus shown that $B$ has trivial null space, so $B$ is invertible.  (Note that for this last conclusion, we use the fact that $\mathbb{F}_2$ is a field.)
This implies that $\det(B) = 1 + 2\mathbb{Z}$.  Therefore, by the functoriality of the determinant, this implies that $\det(A)$ is an odd integer; so $A$ is invertible as an element of $M_{50}(\mathbb{R})$.
