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Seeking some help with the following midterm review question:

Give a linear transformation $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $T\begin{bmatrix} 1 \\1 \end{bmatrix}=3$ and $T\begin{bmatrix} -1 \\2 \end{bmatrix}=6$ and find a matrix $A$ such that $T=T_A$. So $T$ is left multiplication by $A$.

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closed as off-topic by user137731, Claude Leibovici, user26857, Shailesh, Namaste Oct 30 '16 at 11:40

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  • $\begingroup$ I think the only was this can work is to pick a 1 by 2 matrix $[a ..b]$. If you perform the multiplications, you get a system of two linear equations with 2 unknowns from which $a$ and $b$ can be found. Otherwise I wouldn't know... $\endgroup$ – imranfat Oct 30 '16 at 2:26
  • $\begingroup$ @ levap thank you for your edit. Any idea how to proceed with thus problem? $\endgroup$ – jh123 Oct 30 '16 at 3:09
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You can substitute $A = \begin{bmatrix} a & b\end{bmatrix}$ and solve the resulting system of equations. Alternatively:

$$A \begin{bmatrix} 1&-1\\1&2\end{bmatrix} = \begin{bmatrix}3&6\end{bmatrix}$$

$$A = \begin{bmatrix}3&6\end{bmatrix}\begin{bmatrix} 1&-1\\1&2\end{bmatrix}^{-1}$$

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  • $\begingroup$ is this the complete answer? $\endgroup$ – jh123 Oct 30 '16 at 3:08
  • $\begingroup$ No, but it should help you figure out the answer. $\endgroup$ – Michael Biro Oct 30 '16 at 3:09
  • $\begingroup$ okay I will try to proceed fromhere $\endgroup$ – jh123 Oct 30 '16 at 3:10
  • $\begingroup$ @ Michael Biro so all I would need to do is solve the system of equations? $\endgroup$ – jh123 Oct 30 '16 at 3:13
  • $\begingroup$ Yes, that should work. $\endgroup$ – Michael Biro Oct 30 '16 at 3:21

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