Question about linear systems of equations Let $X=\{x_1,\cdots,x_n\}$ be a set of variables in $\mathbb{R}$.
Let $S_1$ be a set of linear equations of the form $a_1 x_1+\cdots+a_n x_n=b$ that are independent.
Let $k_1=|S_1|<n$ where $|S_1|$ denotes the rank of $S_1$ (i.e., the number of independent equations). 
That is, $S_1$ does not contain enough equations to uniquely specify the values of the variables in $X$.
How many other equations are needed to solve the system uniquely? The answer is $n - k_1$. 
Let $M$ be a set of $n-k_1$ equations such that $S_1 \cup M$ is full rank (i.e., the system $S_1 \cup M$ can be solved uniquely). My first question is that how can one find a set $M$?
Let $S_2$ be a set of independent equations such that $|S_2|<n$ too. Now I want to find an $M$ such that both $S_1 \cup M$ and $S_2 \cup M$ are uniquely solvable. How can I find such an $M$? Note that $|M|$ must be $\ge \max(n-k_1,n-k_2)$.
What if we extend the question to $S_1,\cdots,S_m$ such that $S_1\cup M,\cdots,S_m\cup M$ are all uniquely solvable?
A partial answer is also appreciated.

Here is a reasonably easy algorithm if you already know some basic stuff about matrices: 
Look at the coefficients matrix's rows as vectors in $\;\Bbb R^n\;$ :
$$A:=\{ v_1=(a_{11},\ldots,a_{1n})\;,\;v_2=(a_{21},\ldots,a_{2n})\,\ldots,v_k=(a_{k1},\ldots,a_{kn})\}\;\;(\text{where $\,k=k_1\;$ for simplicity of notation)}$$
Since you're given the equations are independent that means $\,A\,$ is a linearly independent set of vectors in $\,\Bbb R^n\;$ . 
Well, now just "simply" complete the set $\,A\,$ to a basis of $\,\Bbb R^n\,$ and that's all...you can do this by taking the matrix
$$\begin{pmatrix}a_{11}&a_{12}&\ldots&a_{1n}\\a_{21}&a_{22}&\ldots&a_{2n}\\\ldots&\ldots&\ldots&\ldots\\a_{k1}&a_{k2}&\ldots&a_{kn}\end{pmatrix}$$
and adding each time a new row $\,(b_1,\ldots,b_n)\,$ . Check whether this matrix is singular or not (for example, reducing this matrix and checking the row you added doesn't become all zeros!), and continue on until you get an $\,n\times n\$ regular matrix and voila!
