<s>[INST] 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 ?
 [/INST] 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.
</s>