I have a linear system S of m equations and n variables (m and n are positive integers). A is the augmented matrix of the system S.
I am trying to find how many solutions there are when the first column of A contains all zeros.
My thinking behind this is there can only be one solution due to the free variables. However, I don't know if this would be correct in all cases.
If there are more variables than equations then you cant solve it for a unique solution, i.e infinite solutions. ex $x+y =0$
If you have n equations and n variables then it is solvable; if the system is linearly dependent you have infinite solutions; linearly independent yields one solution. A zero column in nxn matrix points to linear dependence.
If there are more equations than variables you should have infinite solutions due to linear dependence.
This is all assuming that your system is consistent(no contradicting equations).
In General To examine the number of solutions try checking for linear dependence and independence.
If the first column of the matrix consists of zeros, this means that the coefficient of the first variable is zero in every equation in the system (how did this happen, anyway?). This means that if there is any solution at all, then there must be an infinite number of them, since you can set that first variable to any value at all and still satisfy all of the equations.
Another way to see it is this: If in a LSE $y=Ax$, the first column of your matrix $A$ is zero, the observation vector $y$ contains no information about the first variable in $x$. You can choose it freely, it will not change anything with respect to $y$.
Therefore, you might as well take it out: delete the first entry of $x$ and the first column of $A$ so that you have a new LSE with $m$ equations and $n-1$ unknowns that you can try to solve for the second till last variables in $x$. This doesn't change $y$.