Can you find an invertible submatrix?

Given an $mtimes n$ matrix $mathbf{A}$ $(mgneqq n)$, I’m trying to efficiently find all $ntimes n$ invertible submatrices using a computer. Equivalently, I’m trying to find which rows of $mathbf{A}$ can form a linearly-independent set.

As an approach, I know I can iterate over all $m choose n$ submatrices and compute their determinant, but I am hoping for a more efficient route. I thought about iteratively building up a solution, say starting by considering all solutions that include the first row. However, I don’t know how to compute whether a set of fewer than $n$ vectors in $mathbb{R}^n$ is linearly independent. (If it’s exactly $n$ vectors, you test whether the determinant is zero or not; if it’s more than $n$ vectors, it’s obviously dependent; but if it’s fewer, what do you do? Is there an algorithm?)

I am not sure I understand enough about linear algebra and properties of the null space to know how to compute these subsets efficiently. Any help would be appreciated.

Context #1 : The matrix $mathbf{A}$ is not just any matrix. In my application, it’s the null space of a particular matrix $mathbf{B}$. So I am actually looking for the invertible submatrices $mathrm{ null}(mathbf{B})$.

Context #2: If it helps, I am working with a specific type of matrix useful for solving "number-tower puzzles". The objective of the number-tower puzzle is to fill in numbers in a given triangular structure so that every number is equal to the sum of the two numbers below it. Some numbers are specified in advance, providing constraint. Example:

I wondered how many blocks (and which blocks) you had to specify in advance in order for the problem to be uniquely solvable. Because we’re dealing with a sum constraint, we can express this as a matrix property. And because the $n$ nodes in the bottom row uniquely determine the solution, we know that it will always take exactly $n$ independent entries to uniquely determine the solution. But which of these blocks are independent?

A tower with $n$ blocks in the base has $n(n+1)/2$ total blocks. We can form an "ancestral path" matrix $mathbf{A}$ with $n(n+1)/2$ rows and $n$ columns. Entry $i,j$ is the number of upward paths to node $i$ from the base node $j$. (Equivalently, the $j$th column of $mathbf{A}$ lists the values of all the blocks in the pyramid when the $j$th base block has value 1, and all the other base blocks have value 0— a kind of basis set.) If you form a submatrix $mathbf{D}$ by extracting $n$ rows of $mathbf{A}$ (corresponding to $n$ blocks in the pyramid), those $n$ blocks may be filled out independently if and only if $mathbf{D}$ is invertible. The question is then— how do you find all possible such submatrices?

Perhaps the triangular form of the tower offers some simplification, but I’m not sure.