Consider $$A=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 3\\ 3 & 6 & a \end{pmatrix} $$
a) Wich is the range of $A$ in fuction of $a\in \mathbb{R}$?
b) What terms have to satisfy $b \in \mathbb{R}^{3}$ for the equation system $Ax=b$ has solution, with $x \in \mathbb{R}^{3}$ ? How it depends on $a\in \mathbb{R}$?
For a) I did elemental operations and we have that the equivalent matrix is $$R=\begin{pmatrix} 1 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & a-9 \end{pmatrix}$$ but it doesn't matter wich is the value of $a$ because $\begin{pmatrix} 0 & 0 & a-9 \end{pmatrix}= (a-9)\begin{pmatrix} 0 & 0 &1 \end{pmatrix}$, so the $range(A)=2$ always , or Am I wrong? Maybe I didn't uderstand the problem.
Hint: The range is a subspace of $\Bbb R^3$. So for each $a$, we get a two-dimensional subspace, $V_a=\{(x+3y,2x+3y,3x+ay):x,y\in\Bbb R\}$. (The dimension is two because that is the rank.)
On $b)$, we have that $b$ must be in the range, which is the answer from $a)$. Thus we need $b=(x+3y,2x+3y,3x+ay)$ for some $x,y\in\Bbb R$. Say $b=(b_1,b_2,b_3)$. Then we can get $x$ and $y$ in terms of $b_1$ and $b_2$. We have $\begin{pmatrix}1&3\\2&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}b_1\\b_2\end{pmatrix}$. You can solve for this because the matrix is invertible, and then get $b_3$ in terms of $b_1,b_2$ and $a$ (since $b_3=3x+ay$).