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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} $??

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  • $\begingroup$ Welcome to math.SE, Sofia! I've edited your post in order to make it more readable. Please, visit the help page, to learn how this site works and also how to typeset your question with $\LaTeX$! =) $\endgroup$ – Andrea Orta Jul 29 '13 at 12:39
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What you now have to do is solve the system of equations $$x_1-3x_2+2x_3=1$$ $$x_2+2x_3=3$$

What happens when you solve for $x_2$ in the second equation? Hint: (use a parameter, like let $x_3=t$)

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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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