Let $A$ be a $4 \times 3$ matrix and let $\mathbf{b} \in \mathbb{R}^4$. How many solutions could the system $A \mathbf{x} = \mathbf{b}$ have if $N (A) = \{ \mathbf{0} \}$? Answer the same question if $N(A) \neq \{ \mathbf{0} \}$. Explain your answers.
My Attempt:
** If the nullspace of $A$ is trivial, then the echelon form of the coefficient matrix $A$, and hence the echelon form of the augmented matrix $[ A | \mathbf{b} ]$, of the given system have the full three pivots. Hence the reduced echelon form of the augmented matrix will be of the form $$ \left[ \begin{matrix} 1 & 0 & 0 & c_1 \\ 0 & 1 & 0 & c_2 \\ 0 & 0 & 1 & c_3 \\ 0 & 0 & 0 & c_4 \end{matrix} \right]. $$ So if $N(A) = \{ \mathbf{0} \}$, then the given system has no soluion (if $c_4 \neq 0$) or a unique solution (if $c_4 = 0$).
On the other hand, if $N(A) \neq \{ \mathbf{0} \}$, then the reduced echelon form of the augmented matrix will be a matrix of the form $$ \left[ \begin{matrix} 1 & t_1 & t_2 & c_1 \\ 0 & 0 & 0 & c_2 \\ 0 & 0 & 0 & c_3 \\ 0 & 0 & 0 & c_4 \end{matrix} \right] $$ or $$ \left[ \begin{matrix} 1 & 0 & t_1 & c_1 \\ 0 & 1 & t_2 & c_2 \\ 0 & 0 & 0 & c_3 \\ 0 & 0 & 0 & c_4 \end{matrix} \right] $$ In either case our system has either no solution or infinitely many solutions. **
Is my reasoning correct? If so, have I managed to arrive at the right conclusions as well in both cases? Or, are there any problems?