When $m=n$, we get unique solution that is trivial and hence linearly dependent. So, we get $0$ linearly independent solution (Here, $n-m=0$).
When $m<n$, we also get non-trivial solution along with the trivial solution. We discard the trivial solution because we want only linearly independent solutions. Since we have $m$ linearly independent rows, the rank of $m \times n$ matrix is $m$. Hence $n-m$ linearly independent solutions
The number of linearly independent rows is equal to number of linearly independent columns. Thus, number of linearly independent rows cannot exceed the number of columns. So, $m>n$ is an impossible case.
The correct option is $(C)$ $n-m$.
(I edit my text for providing a solution closer to what you are looking for).
The fact that you give the general solution of the original system has no correlation with the question "solve the corresponding homogeneous system".
[In fact, it will be important to have both for an inevitable second question which is "deduce the general solution of the homogeneous system". See last line of this answer].
Some keypoints:
a) Consider the issue as looking for the kernel of a $3 \times 4$ matrix $A$ which acts as a linear operator with source space $\mathbb{R}^4$ and range space $\mathbb{R}^3$.
b) Minor the dimension of the range space $dim(Im(A)) \geq 2$ because the first two columns of $A$ are independant.
c) Use the rank-nullity theorem:
$dim(Ker(A))+dim(Im(A)=dim(source \ space))=4$.
Thus $dim(Ker(A)) \leq 2$ and in fact, we are able to exhibit two independent vectors of the kernel (by looking for null linear combinations of the columns of $A$, by trial and error for example, which usually does not take a long time when the coefficients are integers or almost integers as is the case here. The coefficients of the null combinations are naturaly the coordinates of elements of the kernel).
One of the elements of the kernel is $(10,-6,1,0)$, another one, non proportional to the first, is $(0,3,2,-5)$ ; thus we have a basis of the kernel.
As a consequence, a final solution to your question is
$(w,x,y,z)=a(10,-6,1,0)+b(0,3,2,-5)$ for any $a,b$.
Remark: this solution could appear under very different forms according to the basis chosen for $Ker(A)$.
and, consequently, the general solution is
$(w,x,y,z)=a(10,-6,1,0)+b(0,3,2,-5)+(0,2,3,-1)$ for any $a,b$.
Best Answer
You add a parameter for every column without a pivot in the REF form of the coefficient matrix. That is, there are the same number of parameters as the dimension of the null space of $A$. What is the relationship between $\dim NS(A)$ and $\text{rk}(A)$?
Mouse over for relationship...