what can you conclude about the number of solutions of the linear system Ax = b? Let $A$ be a $3\times 4$ matrix. If $$\bf{b = a_1 + a_2 + a_3 + a_4}$$ then what can you conclude about the number of solutions of the linear system $A\bf{x = b}$? Explain.
How to solve this? I am trying but did not get anywhere.
Help please.
 A: Note that $A$ is a linear map from $\mathbb{R}^{4}$ to $\mathbb{R}^{3}$. If there is no restriction to the vectors $a_{1},...,a_{4}$ then we can not say anything about the solutions. Could be none or could be many.
If $a_{1},...,a_{4}$ here are the column vectors of $A$‚ then recall that the column vectors of $A$ span the range of $A$. So if $b$ is a linear combination of the column vectors of $A$ then $b$ is in the range of $A$. Hence there is at least one solution, call it $x$.
Now note that by rank nullity theorem $4=\mathrm{null}(A)+$rank$(A)$. Here rank$(A)$ is bounded from above by $3$ so $A$ must have a non-empty kernel. Now $x+e$ is another solution for any $e\in$Ker$(A)$. In other words, there are infinitely many solutions. 
A: If ai for I=1..4 are the columns then one particular solution is x=(1,1,1,1) as you can easily check to which you can ad the arbitrary vector (I-GA)h for any h and get the whole solution set. For an arbitrary y just  take x=Gy.
Note that G is right inverse of A, such that AG=I (for example G=A'*inv[AA'] if rank(A)=3) and thus AGA=A and (I-GA)h is the orthogonal projection of h into ker(A). 
