Find a vector $\mathbf x$ whose image under $T$ is $b$.

Question

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} $??

Answer 1

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 $x_3=t$)

Answer 2

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 $b$.