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If $ \ \ T \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}x_2\\x_1 - x_2\\2x_2 + x_3\end{bmatrix}$ find $T^{-1}$.

How can do I this? I know how to find the inverse of any matrix, you just combine it with its Identity Matrix and then you row reducing to find the inverse matrix, but I am not sure what to do here.

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    $\begingroup$ Why not start by finding the matrix of $T$ with respect to the standard basis? $\endgroup$ – carmichael561 Jul 5 '16 at 0:36
  • $\begingroup$ Oh, good point @carmichael561 $\endgroup$ – Shammy Jul 5 '16 at 0:37
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HINT: One straightforward way is to write down $T$ and then use any standard method to find its inverse. For instance, it’s clear that the first row of $T$ must be $\begin{bmatrix}0&1&0\end{bmatrix}$; why? You should be able to work out the other two, but I’ll leave them spoiler-protected below.

$\begin{bmatrix}1&-1&0\end{bmatrix}$ and $\begin{bmatrix}0&2&1\end{bmatrix}$

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  • $\begingroup$ so the answer should be $T^{-1} = \begin{bmatrix}0&1&0\\1&-1&0\\0&2&1\end{bmatrix}$ correct? $\endgroup$ – Shammy Jul 5 '16 at 0:40
  • $\begingroup$ @Shammy: No, the matrix that you have there is $T$; now you need its inverse. $\endgroup$ – Brian M. Scott Jul 5 '16 at 0:41
  • $\begingroup$ So should I now row reduce $T^{-1} = \begin{bmatrix}0&1&0&1&0&0\\1&-1&0&0&1&0\\0&2&1&0&0&1\end{bmatrix}$ $\endgroup$ – Shammy Jul 5 '16 at 0:44
  • $\begingroup$ @Shammy: Yes, but $T^{-1}$ won’t be the whole row reduced matrix: it will be the righthand half. $\endgroup$ – Brian M. Scott Jul 5 '16 at 0:46
  • $\begingroup$ Thank you! I got \begin{bmatrix}1&1&0\\1&0&0\\-2&0&1\end{bmatrix}$ $\endgroup$ – Shammy Jul 5 '16 at 0:55
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hint: $T^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 2 & 1 \\ \end{bmatrix}^{-1}$

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  • $\begingroup$ great too @DeepSea ty $\endgroup$ – Shammy Jul 5 '16 at 0:57
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This is the same problem as

If $ \ \ S \begin{bmatrix}x_2\\x_1 - x_2\\2x_2 + x_3\end{bmatrix} = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} $ find $S$.

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