Prove a property of a determinant with integer entries with one variable We have a square matrix $A$ of size $2n\times 2n$ (where $n>2$) with entries from the set $\{-1,0,1,t,-t\}$. Such that:
1) matrix $A$ has $n$ pairs of rows (in the picture a pair is bordered blue), where one (upper) row from a pair has exactly two nonzero entries $-t, t$. The second row from the pair has exactly four nonzero entries $1, 1,-1,-1$ such that below $-t$ is $0$, and below $t$ is $1$.
2) in every column of $A$ we have exactly two elements from the set $\{-1,1\}$.
We construct now a square matrix $B$ of size $(2n-1)\times(2n-1)$ from $A$ be removing the last row and column. It is clear that $det(B)$ is a polynomial (of variable $t$) of degree at most $n$.
Prove that if $deg(det(B))=n$, then the leading coefficient is $1$ or $-1$.

 A: We will use column operations to reduce the matrix into an upper triangular form and then we can calculate the derivative easily by examining the pivots. We'll proceed by row pair, starting at the top.
So to do this we first swap columns to put the $-t$ entry on the diagonal and we pair it with the $t$ on the upper off-diagonal. This forms a $2 \times 2$ block matrix of the form $$\pmatrix{-t & t \\ 0 & 1}$$ Now  add the first column to the second which leaves the determinant unchanged and now this block, call it $A$, is in the form $$ A = \pmatrix{-t & 0 \\ 0 & 1}$$
There are two points to be made now. Firstly, there may be other stuff below these two entries but we're looking to make it lower triangular so we're only interested in what happens above the diagonal. But now since the rest of the columns in the row pair to the right of $A$ are of the form $\pmatrix{ 0 \\ \pm 1}$ we can add or subtract the second column of $A$ to these other columns to zero them out above the diagonal.
Edit: To do - prove that with columns having more than one $t$ term that zeroing out the $\pm 1$ terms above the diagonal will leave the diagonal term equal to $\pm t$ or $0$ using that there are exactly four non-zero entries $-1,-1,1,1$. 
It's possible that zeroing out will leave a $2$ or $0$ on the diagonal. In the $2$ case, move the column that would contain the offending $2$ between the columns of $A$ and add it to the second column of $A$, zeroing it out above the diagonal instead. Note that in this case, because of the restriction that each column must contain exactly two $\pm 1$ and that there are no further complications from adding these two vectors. By moving the column we may now have put a $\pm 1$ or $0$ on the diagonal which can make the determinant zero or change the sign.
Now that we have the top row pair in the form we like, we can restrict our attention to the $2(n-1) - 1 \times 2(n-1)-1$ submatrix that excludes the top row pair and the first two columns of that row pair and repeat until the matrix is upper triangular.
In this form, we can see that either the diagonal is alternating pairs of $-t$ and $\pm 1$, but it's also possible one of them will now be zero. Also, deleting the last row and column removes either a $0$ or a $1$, but never a $-t$ because of how we reduced it.
Since we know the determinant has rank $n$ we can't have a zero on the diagonal and there must be $n$ entries that are $-t$. Since they are alternating along the diagonal that has $2n-1$ then $n-1$ of them will be $1$ and the other $n$ will be $-t$ so the determinant is exactly $(-t)^n  = (-1)^nt^n$ and so the leading coefficient is $\pm 1$ after accounting for the sign change due to the column swapping.
