My Math Forum Solve for x in T(x) = Ax in linear transformation

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February 10th, 2019, 09:54 PM   #1
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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!
Attached Images
 qn.jpg (61.3 KB, 9 views) error.jpg (4.3 KB, 4 views)

 February 11th, 2019, 03:54 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 The calculator's answer is correct and unique. Can you post your detailed working? Thanks from topsquark and Appletree
February 11th, 2019, 05:01 PM   #3
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Joined: Apr 2016
From: Wonderland

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After re-writing 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 re-written 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?
Attached Images
 working.jpg (79.2 KB, 3 views)

 February 11th, 2019, 06:16 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 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: 20,966 Thanks: 2216 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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