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i am following this guide: http://www.calpoly.edu/~brichert/teaching/oldclass/f2002217/handouts/goof.pdf

my question is to find the linaer transformation that adheres to

$T(1,1,1) = (1,1,1)$

$T(0,1,0) = (0,1,0)$

$T(1,0,2) = (1,0,1)$

my matrix looks like this:

$\begin{pmatrix} 1 & 1 & 1 & | &b_1\\ 0 & 1 & 0 & | & b_2 \\ 1 & 0 & 2 & | & b_3 \\ \end{pmatrix}$

after reduction it looks like this:

$\begin{pmatrix} 1 & 0 & 0 & | &2b_1 -2b_2 -b_3\\ 0 & 1 & 0 & | & b_2 \\ 0 & 0 & 1 & | & b_3 +b_2 -b_1 \\ \end{pmatrix}$

i am getting that:

$T(b_1,b_2,b_3) = (b_1-b_2,2b_1-b_2-b_3,b_1-b_2)$

this is not the actual solution, it should be $(b_1,b_2,b_1)$

what am i doing wrong, i believe i havent made any arithmetic errors

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    $\begingroup$ the "actual solution" is also wrong, it contradicts $T(1, 0, 2) = (1, 1, 1)$. $\endgroup$
    – hunter
    Jan 20 '14 at 13:34
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    $\begingroup$ the matrix reduction is also wrong, first row should be $2b_1 - b_2 - b_3$ $\endgroup$
    – hunter
    Jan 20 '14 at 13:35
  • $\begingroup$ made a typo in the original question, its fixed now $\endgroup$
    – guynaa
    Jan 20 '14 at 13:36
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Your matrix must be:

$\begin{pmatrix} 1 & 0 & 1 & | &b_1\\ 1 & 1 & 0 & | & b_2 \\ 1 & 0 & 2 & | & b_3 \\ \end{pmatrix}$

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  • $\begingroup$ but... in the link above the vectors are put in rows! $\endgroup$
    – guynaa
    Jan 20 '14 at 13:43
  • $\begingroup$ It seems there is a typo in the link. $\endgroup$ Jan 20 '14 at 13:49
  • $\begingroup$ that solved it, thanks! $\endgroup$
    – guynaa
    Jan 20 '14 at 13:52

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