Parametrized linear operator I've been trying to solve the following task:

Determine $a$, $b$ $\in \mathbb{R}$ so that for the linear mapping $A :\mathbb{R}^3\to\mathbb{R}^3 $, with linear transformation matrix $$\mathcal{M}(A)=\begin{bmatrix}
    a       & 1 & 1\\
    1       & b & 1 \\
    1       & 2b & 1
\end{bmatrix}$$ holds that in a standard basis vector $v=(4,3,4) \in \mbox{null}(A) $

This seemed pretty straightforward to me: I just multiply $\mathcal{M}(A)$ with $v^T$, equate that with zero and just calculate the parameters, i.e.,
$$4a+7=0\\8+3b=0\\8+6b=0$$
This system of course, has no solution and the textbook explicitly states that the solution exists ( maybe the author predicted someone would make this error ).
I believe that the problem with my solutions is : $\mathcal{M}(A)$ is not necessarily a linear operator matrix with respect to standard base. Instead, this is some other base and I'd have to change the base in order to get the correct result. Is  this true? How should I got about doing this?  

the problem has a solution in the sense of determining a matrix
$N$ similar to $M(A)$ such that the vector $v=(4,3,4)$ is in the $null(N)$. in
this case, just take $N =  P^{-1}MP$, where $P =\left(
\begin{array}{ccc}
 -\frac14 & 0 &0\\
3 & -4& 0\\
0&-1&1
\end{array}
\right)$ then $Nv^T = (0,0,0)$ for all $b\in\Bbb{R}$ with $a = 1$
