$\quad$ Let $A$ be a $5$x$4$ matrix.There exists a vector(column) $B=[0,1,2,3,4]^{T}$ such that the solution space of $AX=B$ is $[1+2s,2+3s,3+4s,4+5s]^{T}$where $s \in \mathbb R$. Find the rank of the matrix.
$a)1 \quad b)2 \quad c)3 \quad d)4$
Arguments:
since its a $5 \times 4$ matrix, the number of unknowns is less than the number of equations, so the maximum rank can be $4$.
since $s \in \mathbb{R}$ the solution space is infinite, we have a result that
"System of m linear equations in n unknowns has a unique solution iff the rank of the matrix $A$ = rank of the augmented matrix $[A;B] =n$"
Now, the rank cannot be $4$, since the number of unknowns is $4$ and we have infinite number of solutions, it cannot be $4$ if it is then we must have a unique solution.so the rank can be $3$ or $2$ or $1$.
I don't know about the other $3$ options.
Many people said that
"since $s$ is the only variable, the nullity is $1$ and hence the rank is $3$. Is it true? Nullity is the dimension when $AX=0$ and not $AX=B$" Am I right?
Please do correct me where I am wrong..
The space of solutions of an inhomogeneous system of equations $Ax = b$ (where $b \neq \vec{0}$) is not a linear subspace but an affine subspace whose dimension (as an affine subspace) is the same as the dimension of the space of solutions for $Ax = \vec{0}$ (which is $\dim \ker {A}$).
If you are not familiar with affine subspaces, note that the space of solutions for $Ax = b$ can be always written as $x_0 + \ker A$ where $x_0$ is some (arbitrary) particular solution satisfying $Ax_0 = b$. In your case, you have
$$ \{ x \, | Ax = b \} = [1,2,3,4]^T + \mathrm{span} \{ [2,3,4,5]^T \}. $$
Check that this implies that $\ker A = \mathrm{span} \{ [2,3,4,5]^T \}$ and thus $\mathrm{rank} A = 3$.