Is this proof correct?
Proof: Let $a_{1},a_{2},...,a_{n}$ denote the columns of the matrix with one row eliminated. The sum of the entries in each row of the matrix is $t+(1-t)+(-1)=0$. Thus $a_{1}+a_{2}+...+a_{n}$ is a zero vector. Suppose some column $a_{p}$ has been eliminated but we want to eliminate column $a_{q}$ instead. As $a_{1}+a_{2}+...+a_{n}=0$, $a_{q}=-a_{p}-a_{1}-a_{2}-...-a_{q-1}-a_{q+1}-...-a_{n}$. After we eliminated $a_{q}$ we are left with some $(n-1)\times(n-1)$ submatrix $M$. Also, let the submatrix we got by eliminating $a_{p}$ be $L$. We know that $\text{det}(L)=\Delta_{k}(t)$, the Alexander polynomial. The only real difference between matrices $L$ and $M$ is one has $a_q$ but no $a_{p}$ and vice versa. However, by some column operations we can change the $a_{p}$ in $M$ into $a_{q}$. After some column interchanges, the matrix $M$ will arrive to matrix $L$. These operations only change the determinant of $M$ by a factor of $+1$ or $-1$. So, the Alexander polynomial does not depend on the column eliminated as $\text{det}(L)=\pm\text{det}(M)$