Let $A$ be a matrix over $\mathbb{Z}$ of the following form.
$$A= \begin{bmatrix} 1 & 0 & \cdots & 0\\ 0 & a_{22} & \cdots & a_{2n}\\ \vdots & \cdots & \ddots & \vdots\\ 0 & a_{m2} & \cdots & a_{mn} \end{bmatrix}=\begin{bmatrix} 1 & \mathbf{0}\\\mathbf{0} & B\end{bmatrix}.$$
Claim 1. If Smith normal form of $B$ is has invariant factors $d_2,\cdots, d_k, 0 ,\cdots$, then for $A$, the invariant factors are $1,d_2,\cdots, d_k, 0 ,\cdots$.
Proof: This is simply because if $P$ and $Q$ are invertible matrices over $\mathbb{Z}$ such that $PBQ$ is in Smith normal form with invariant factors $d_2,\cdots, d_k, 0 ,\cdots$ then, the product $$ \begin{bmatrix}1 & \mathbf{0}\\\mathbf{0} & P\end{bmatrix} A \begin{bmatrix}1 & \mathbf{0}\\\mathbf{0} & Q\end{bmatrix} $$ will be almost diagonal with entries $1,d_2,\cdots, d_k, 0,\cdots, 0$ and these entries satisfy the codition of invariants in Smith normal form (i.e. $1|d_2$, $d_2|d_3$, $\cdots$) so by uniqueness of Smith normal form, we can conclde that above product gives Smith normal form of $A$ with invariant factors $1,d_2,\cdots, d_k,0,\cdots$.
Claim 2. GCD of $2\times 2$ minors of $A$ is equal to GCD of $1\times 1$ minors of $B$.
Proof: If previous claim is correct, then $d_2=1.d_2$ is GCD of $2\times 2$ minors of $A$.
Question 1. Are the above claims with proof are correct?
Question 2. What are some simple methods (tricks) to compute Smith normal form (other than finding GCD's of minors), which can be implemented nicely in computation, rather than going by first principle?