Determining the dimension of the set of solutions of the system $Ax=b$ without solving for $x$ or applying row reduction

Question

Consider the matrix $A=\begin{bmatrix}2&2&2&4\\1&2&0&-1\\1&3&-1&-4\end{bmatrix}$

part d) What is the dimension of the solution space of the homogeneous system $Ax=0$ and the dimension of the set of solutions (general solution) of the system $Ax=b$ (Do not solve for $x$ or apply row reduction).

Well, the dimension of the solution space of the homogeneous system $Ax=0$ means the dimension of the null space of A, which is $nullity(A)$. I found $rank(A)$ as $2$ in one of the previous parts. $2 + nullity(A) = 4$, $nullity(A) = 2$. But, what will I do about the general solution?

(No $b$ given.)

Answer 1

The dimension of the general solution of $Ax=b$ is the same as the dimension of the
solution of the homogeneous equation $Ax=0$ because the general solution of the equation $Ax=b$ is the sum of the general solution of the equation $Ax=0$ and a special solution of $Ax=b$. If the system is unsolveable, then the dimension is, of course, $0$.

Answer 2

Singular value decomposition is often the best way of finding the rank assuming you have SVD libraries close at hand.