Condition for solution of linear system The problem is:

Let $A$ be a singular $N \times N$ matrix with real entries and let $\mathbf{b}$ be a $N \times 1$ matrix (i.e. column vector). Let $A^\top $  and $\mathbf{b}^\top$  be their respective transposes. Consider the system of $N$ linear equations in $N$ unknowns written in the matrix form as:
  $$                                                                                                          A^\top\mathbf{x} = \mathbf{b}.
$$
  Complete the following sentence: The above linear system has a solution, if and only if, $$\mathbf{b}^\top \mathbf{u} = 0,$$ for all column vectors $\mathbf{u}$ such that …

Please help me in filling up the above blank.Thank you in advance.
 A: You know that $A^Tx = b$ has a solution only if $b$ is in the space spanned by the columns of $A^T$ (which are the rows of $A$). So you could say that $b$ must be orthogonal (ie scalar product is $0$) to the orthogonal space of $\text{Col } A^T$
(So if $b$ is orthogonal to the orthogonal of $\text{Col }A^T$ it means that it must be in $\text{Col }A^T$, which is what we want)
Now, the orthogonal complement of $\text{Col }A^T$ is $\text{Ker } A$! So in the end you can say that for every $u \in \text{Ker } A$ it must hold that 
$$b^Tu = 0$$
@A.Chattopadhyay. Is that your comment or are you reporting what your professor told you? Anyhow, you must have misunderstood what I wrote; yours is not a countexample. In fact, if $A^T = \begin{pmatrix}1 & 0 &  1 \\ 0 & 1  & 1 \\  1& 1 &2\end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ -1 \\ 0\end{pmatrix}$, we have that $\text{ Ker } A  = \left\{u \in \mathbb R^3: u = \begin{pmatrix} -t \\ -t \\ t\end{pmatrix} \text{ for } t \in \mathbb R\right\}$. Hence for every $u \in \text{Ker } A$ we have $$u^Tb = -t\cdot 1 -t\cdot(-1) + t\cdot 0 = 0$$ as expected
