Solution Space - Linear Algebra For a matrix:\begin{bmatrix}-1&2&3&-3&6&7\\
1&-1&-2&2&-5&-6\\
-1&1&2&-1&2&4\\
-2&2&4&-2&4&8\\\end{bmatrix}
To solve for whether a specific vector  is inside the solution space, you take the reduced-row echelon form and times it by the vector. If it equals 0, it is in the solutionspace.
Can you please explain why this is the case?
 A: I'm not 100% certain I fully understand the question, but I believe I have the general idea. (Correct me if I'm wrong). 
I take it by solution space you're referring to the set of vectors that map to the kernel of this matrix, or - in other words - are solutions to the homogeneous case of this matrix:
\begin{bmatrix}-1&2&3&-3&6&7 & | \space0\\
1&-1&-2&2&-5&-6 & | \space0\\
-1&1&2&-1&2&4 & | \space0\\
-2&2&4&-2&4&8 & | \space0\\\end{bmatrix}
There are a few ways to go about this. 
(1) Simply multiply by your vector [4 -3 2 -1 -1 1]^T. If the 0 vector is the product, then it's in the solution space.
(2) Reduce the matrix to row echelon form, and multiply it again. Again, if the 0 vector is the product, then it's in the solution space.
Note that (1) & (2) are equivalent. The only difference is that you reduced it. 
However, if you want to formally prove that the vector is in the solution space, which is what I think you're asking, then you need to use the row echelon form of the matrix. (Note that 1 and 2 are sufficient ways to show that the vector is in the solution space).
The row echelon form of this matrix is 
\begin{bmatrix}1&0&-1&0&-1&-3\\
0&1&1&0&-2&-1\\
0&0&0&1&-3&-2\\
0&0&0&0&0&0\\\end{bmatrix}
As you can see, the matrix has rank 3, and therefore has 3 free variables. From here you can write a basis for the null space and construct a matrix composed of the basis vectors. Finally, just show that the vector you mentioned is in the range of this matrix.
A: First of all, if you multiply the matrix by a vector and you get zero, the vector is in the solution space by definition.
Now, if you multiply the vector by a reduced form of the matrix, the vector is in the solution space.
Go back to the motivation of a reducing a matrix. Whenever you reduce a matrix, you are solving a system like here or here.
