<s>[INST] Find a vector $\mathbf x$ whose image under $T$ is $b$. I am having trouble with this question and how to get the answer.
With $T$ defined by $T(\mathbf x)=A\mathbf x$, find a vector $x$ whose image under $T$ is $b$.
$$
A =  \begin{pmatrix}
1 & -3 & 2  \\ 
3 & -8 & 8  \\
0 & 1 & 2  \\
1 & 0 & 8
\end{pmatrix} \qquad,\qquad
b = \begin{pmatrix} 
1  \\
6  \\
3  \\
10
\end{pmatrix}
$$
What I have done so far is that I've combined the two matrices into a augmented matrix. And row reduced it to get:
$$\begin{pmatrix}
1 & -3 & 2 & 1 \\  
0 & 1 & 2 & 3 \\  
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
So does this just mean that the answer to the question is $\mathbf x = \begin{pmatrix} 
1  \\ 3  \\ 0  \\ 0 \end{pmatrix} $??
 [/INST] Hint 1: If $A$ has $3$ columns, the dimension of $x$ must be $3$.
Hint 2: To check your result, compute $Ax$ and see if you got $b$.
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