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Let $T(a,b)=(a+b,2a-b,3a)$

Find the image of $T$ (as a span of vectors).

So I created the augmented matrix and got this: $A$= $\begin{bmatrix}1 & 1 & b_1\\2 & -1 & b_2\\ 3 &0 & b_3\end{bmatrix}$

then I did -3r1+r3 $A$= $\begin{bmatrix} 1& 1 &b_1\\2 & -1 & b_2\\ 0 &-3 & -3b_1+b_3\end{bmatrix}$

then I did -2r1+r2 $A$= $\begin{bmatrix}1 & 1 & b_1\\0 & -3 & -2b_1+b_2\\ 0 &-3 & -3b_1+b_3 \end{bmatrix}$

Finally -b2+b3 $A$= $\begin{bmatrix}1 & 1 & b_1\\0 & -3 & -2b_1+b_2\\ 0 &0 & -b_1-b_2+b_3\end{bmatrix}$

Now I know in order for the system to be consistent $-b_1-b_2+b_3$ must equal $0$ but how do I find the image of $T$ as a span of vectors?

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  • $\begingroup$ Image of $T$ is spanned by (1,2,3) and (1,-1,0). What are you trying to do with the augmented matrix ? $\endgroup$ Commented Feb 10, 2016 at 1:39
  • $\begingroup$ I'm trying to find the restrictions regarding the b values $\endgroup$ Commented Feb 10, 2016 at 1:40
  • $\begingroup$ But doesn't that answer your question ? What is the need of finding these b values ? $\endgroup$ Commented Feb 10, 2016 at 1:46
  • $\begingroup$ at this point I have b3=b1+b2 but what is the actual span of vectors.. I'm just confused what my final answer would be. Do I solve each of those b's in terms of one variable? $\endgroup$ Commented Feb 10, 2016 at 1:47
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    $\begingroup$ You asked this yesterday and got a comprehensive answer here: math.stackexchange.com/questions/1647004/…. There is no reason for you to row reduce the matrix (unless you suspect some of the vectors in your column space to be dependent on the others). The image is just $AX$ where $X \in F^2$. $\endgroup$ Commented Feb 10, 2016 at 1:58

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$T(a,b) = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}a + \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}b$, so image $T$ $=$ span$\left\{ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right\}$ by definition. No need to solve a system of linear equations for this question.

EDIT

If for some reason you must use row reduction: the condition for \begin{pmatrix} b1 \\ b2 \\ b3 \end{pmatrix} to be in image $T$ is $b3 = b1+b2$, as you calculated. Thus $\begin{pmatrix} b1 \\ b2 \\ b3 \end{pmatrix} = b1 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + b2 \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$, so

\begin{equation} \mbox{image}\ T = \mbox{span}\left\{ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right\}. \end{equation}

You can see relatively easily that both answers are equivalent, by writing each vector as a linear combination of the vectors in the other set, e.g. $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.$

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  • $\begingroup$ my professor told me I needed to find conditions with b for some reason.... not quite sure why though $\endgroup$ Commented Feb 10, 2016 at 1:58
  • $\begingroup$ @Lil OK then, see my edit. $\endgroup$ Commented Feb 10, 2016 at 5:54

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