Image of linear system I found the determinant and null space of the matrix in previous exercises, but I am having trouble understanding how to find the image of following matrix.
"Given the matrix
$$A_{a} = 
 \begin{pmatrix}
  a & 1 & 1 \\
  a & 0 & a \\
  0 & 1 & a
 \end{pmatrix}$$
for $a\in \mathbb{R}$, determine the image
$$B(A_{a})=\{y\in \mathbb{R}^3 \mid A_ax=y\;,\;x\in \mathbb{R}^3 \}"$$
To me, it seems the equation $A_ax=y$ have so many unknowns; a, x and y?
Solutions and hints are very welcome.

A general method is with Gaussian elimination. We can do
\begin{align}
\begin{pmatrix}
  a & 1 & 1 \\
  a & 0 & a \\
  0 & 1 & a
\end{pmatrix}
&\to
\begin{pmatrix}
  a & 1 & 1 \\
  0 & -1 & a-1 \\
  0 & 1 & a
\end{pmatrix} && R_2\gets R_2-R_1
\\
&\to
\begin{pmatrix}
  a & 1 & 1 \\
  0 & -1 & a-1 \\
  0 & 0 & 2a-1
\end{pmatrix} && R_3\gets R_3+R_2
\\
\end{align}


*

*If $a\ne0$ and $a\ne 1/2$, the rank of the matrix is $3$, so the image is $\mathbb{R}^3$.

*If $a=1/2$, the rank is $2$ and the image is generated by the first two columns of the matrix, so a basis consists of
$$
\begin{pmatrix}1/2 \\ 1/2 \\ 0\end{pmatrix},
\qquad
\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}
$$

*If $a=0$, the rank is $2$ and the image is generated by the last two columns of the matrix, so a basis consist of
$$
\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix},
\qquad
\begin{pmatrix}1 \\ 0 \\0\end{pmatrix}
$$

