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Let T be a linear transformation from $R^3$ to $R$ such that $T(1,1,1) = 1$, $T(1,1,0) = 2$ and $T(1,0,0) = 3$. Find $T(0,1,1)$. I'm trying to solve it by this way: $R1 - R3$ -> $R1$. Is it correct? $$ \left[ \begin{array}{ccc|c} 1&1&1&1\\ 1&1&0&2\\ 1&0&0&3\\ \end{array} \right] $$ =>$$ \left[ \begin{array}{ccc|c} 0&1&1&-2\\ 1&1&0&2\\ 1&0&0&3\\ \end{array} \right] $$

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closed as off-topic by user223391, Did, Dando18, user284331, Trevor Gunn May 9 '18 at 3:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Did, Dando18, user284331, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ $T$ is linear. What does that mean to you? $\endgroup$ – Arthur May 8 '18 at 14:19
  • $\begingroup$ Use matrix to solve? $\endgroup$ – Vinh Quang Tran May 8 '18 at 14:23
  • $\begingroup$ What is your try. Please add you try to the question. $\endgroup$ – Amin235 May 8 '18 at 14:26
  • $\begingroup$ Matrices are part of the theory around linear transformations, yes. And this can probably be solved with matrices. But that's not what I asked you about. I did not ask what consequences "$T$ is linear" has on how one would solve the problem. That is the next step. I asked what you think "linear transformation" means. Without that you can't even begin to solve this problem. $\endgroup$ – Arthur May 8 '18 at 14:28
  • $\begingroup$ I added my try to the question. $\endgroup$ – Vinh Quang Tran May 8 '18 at 14:31
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Since $$ (0,1,1) = (1,1,1) - (1,0,0), $$ we have \begin{align} T(0,1,1) &= T((1,1,1) - (1,0,0))\\ &= T(1,1,1) - T(1,0,0)\\ &= 1 - 3\\ &= -2. \end{align}

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  • $\begingroup$ This is wrong. You need to take another look at $-T(1,0,0)$. $\endgroup$ – Arthur May 8 '18 at 14:50
  • $\begingroup$ @Arthur Thanks, corrected the typo. $\endgroup$ – Math1000 May 8 '18 at 15:06
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    $\begingroup$ Actually this is the right way to address these plz-do-hwk-asap questions with no context: to give deliberately a wrong answer. So simple and so effective... $\endgroup$ – Did May 8 '18 at 15:06
  • $\begingroup$ @Did The answer has been corrected... $\endgroup$ – Math1000 May 8 '18 at 16:10

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