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February 10th, 2019, 09:54 PM  #1 
Newbie Joined: Apr 2016 From: Wonderland Posts: 23 Thanks: 0  Solve for x in T(x) = Ax in linear transformation
When solving for x as given in question (photo attached), I can't seem to get the right x. I solve for x using row reduction with augumented matrix 1 4 4 4 0 1 4 1 2 9 8 3 and I get x = (4,3,1) which is wrong, with the error message that I'm solving wrongly. Using graphic calculator's simultaneous equation solver, I get an answer of (20, 5, 1), different from my augumented matrix. I don't understand why, and I'm confused. Where did I go wrong? Any tips is greatly appreciated! 
February 11th, 2019, 03:54 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039 
The calculator's answer is correct and unique. Can you post your detailed working?

February 11th, 2019, 05:01 PM  #3 
Newbie Joined: Apr 2016 From: Wonderland Posts: 23 Thanks: 0  Answer
After rewriting my working out to present here, I realised I made a calculation error in my very first step of augmenting the matrix, which snowballed greatly. I attached my rewritten working which got me the right answer. Edit: I still do not understand the message of "Solve T(x) = b for x. That is A(x) = b." What does it mean? 
February 11th, 2019, 06:16 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
The image of $x$ under the transformation $T$ is written $T(x)$. You are told that the definition of the transformation $T(x)$ is $Ax$: $T(x)=Ax$. So if $T(x)=b$, that means that $Ax=b$.

February 11th, 2019, 11:19 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039 
A$\text{x = b}$ has solution $\text{x}\,=$ A$^{1}\text{b}$, where A$^{1} = \begin{pmatrix} 7 & 1 & 3 \\ 2 & 0 & 1 \\ 1/2 & 1/4 & 1/4 \end{pmatrix}$. 

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linear, linear transformation, solve, transformation 
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