 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$^{-1} = \begin{pmatrix} 7 & 1 & -3 \\ 2 & 0 & -1 \\ 1/2 & -1/4 & -1/4 \end{pmatrix}$. Tags linear, linear transformation, solve, transformation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post evzquz Linear Algebra 1 November 16th, 2016 04:55 AM bilano99 Algebra 2 April 2nd, 2012 11:21 AM remeday86 Linear Algebra 4 July 25th, 2010 12:47 PM tinynerdi Linear Algebra 3 February 2nd, 2010 03:09 AM ypatia Linear Algebra 2 March 5th, 2009 01:49 PM

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