Calculating the Image of $A.$ Where did I err? $$A=\begin{pmatrix}
4 &-1& 1\\
8&-2&2\\
-6&1&-2\\
\end{pmatrix}$$
I have to show $p=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix} \in \mathrm{Im}A=\left\{Ax \mid x\in \mathbb{R^3} \right\}$
If I let $x=\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix},$ then $Ax=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix}=p$ thus $p\in \mathrm{Im}A$.
But when I try to find what  $\mathrm{Im}A$ is,  I'm pazzled.
In order to find Im$A,$ I did elementary transformations for $A$, and I got $$A\to \begin{pmatrix}
1 & 0 & \frac{1}{2} \\
0&1&1\\
0&0&0\\
\end{pmatrix}.$$
And $$\begin{pmatrix}
1 & 0 & \frac{1}{2} \\
0&1&1\\
0&0&0\\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2\\
x_3\
\end{pmatrix}=(x_1+\frac{1}{2}x_3)\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}
+
(x_2+x_3)
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}$$
so Im$A=$span$\left\{
\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix},
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}
 \right\}$
But I cannot write $p$ as a linear combination of $\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}$ and $
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}$
Where did I err ?
 A: Your mistake is that $\mathrm{C\left(A\right) ≠ span}\left(\begin{bmatrix}1\\0\\0\end{bmatrix},\ \begin{bmatrix}0\\1\\0\end{bmatrix}\right)$.
Elimination actually changes the column space of $\mathrm A$, so $\mathrm{C\left(A\right) ≠ C\left(rref\left(A\right)\right)}$.
By reducing $\mathrm A$ you just revealed a number of linearly independent columns, but nothing more.
The actual column space of $\mathrm A$ is the span of vectors of $\mathrm A$, corresponding to pivot columns of $\mathrm{rref\left(A\right)}$.
So in your particular case $\mathrm{C\left(A\right) = span}\left(\begin{bmatrix}4\\8\\-6\end{bmatrix},\ \begin{bmatrix}-1\\-2\\1\end{bmatrix}\right)$.

Just a side note:
It's not usually written as $\mathrm {Im \left(A\right)}$, because the word "image" is referred to a transformation. Meaning, if you have your linear transformation $\mathrm {T\left(\vec x\right) = A\vec x}$, then you can speak of $\mathrm {Im \left(T\right)}$, which is the image of $\mathbb R^n$ under $\mathrm T$ (which is the range of $\mathrm T$). But $\mathrm {Im \left(T\right)}$ is equivalent to $\mathrm {C\left(A\right)}$. They're just the same thing by their definitions.
So it would be more correct to ask about $\mathrm {C\left(A\right)}$ and not $\mathrm{Im \left(A\right)}$ in your case.
A: The image is the span of linearly independent vectors in the column of the matrix. It suffices to show that $p$ is redundant and can be written as a linear combination of the other columns. After your row reduction, notice that only the first two columns have free variables whereas the third doesn't. This automatically implies that the third is redundant and the linear combination should be obvious.
Alternatively, you could have set up the augmented matrix system below:
$$\begin{pmatrix}
4 &-1& | &1 &\\
8&-2& | &2\\
-6&1& | &-2\\
\end{pmatrix}$$
After row reducing, we have,
$$\begin{pmatrix}
1 &0& | &\frac{1}{2} &\\
0&1& | &1\\
0&0& | &0\\
\end{pmatrix}$$
, which is analogous to what you found. In other words, we have that,
$$
\begin{pmatrix}
1 \\
2 \\
-2\\
\end{pmatrix} = \frac{1}{2}\begin{pmatrix}
4 \\
8 \\
-6\\
\end{pmatrix} + \begin{pmatrix}
-1 \\
-2 \\
1\\
\end{pmatrix}$$
, which is easily verified.
