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Find 2 by 3 matrix M such that M $\begin{pmatrix}a \\ b \\ c\end{pmatrix}= \begin{pmatrix}d \\ e\end{pmatrix}$ whenever $(ax^2+bx + c)'=dx+e$.

Here is my take on it. \begin{align*}(ax^2+bx+c)' = 2ax +b = dx+e\end{align*} which gives $a = \frac{d}{2}$ and $b = e$. So we have, \begin{align*}\begin{pmatrix}m_1 & m_2 & m_3 \\ m_4 & m_5 & m_6\end{pmatrix}\begin{pmatrix}a \\ b \\ c\end{pmatrix} = \begin{pmatrix}am_1+bm_2+cm_3\\ am_4 + bm_5 + cm_6 \end{pmatrix}=\begin{pmatrix}2a \\ b\end{pmatrix}\end{align*}After solving the above equation, I got $m_1 +m_4 = 2, m_2 + m_5 = 1$ and $m_3 + m_6 = 0$.

I can just set $m_1, m_2$ and $m_3$ to be free variable but I don't think this is the right answer. And I am not sure if I am going to the right direction either. It will be really helpful for me if someone can point me to the right direction.

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  • $\begingroup$ How'd you get $a=\frac{d}{2}$ and $b=e$ $\endgroup$ – user76568 Jan 31 '18 at 1:41
  • $\begingroup$ $2ax + b = dx + e$ so I figure $2a = d$ and $e = b$. $\endgroup$ – abuchay Jan 31 '18 at 1:42
  • $\begingroup$ You think $x$ is fancier? :) You can't conclude $b=e$ $\endgroup$ – user76568 Jan 31 '18 at 1:45
  • $\begingroup$ In that case, what do you suggest then? Solving for x will just give me, $x = \frac{e-b}{2a-d}$ and I am not sure what to do with it. $\endgroup$ – abuchay Jan 31 '18 at 1:49
  • $\begingroup$ arithmetic and hope $\endgroup$ – user76568 Jan 31 '18 at 1:50
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You have $6$ variables and $2$ constraints. Therefore you will have $4$ free variables... giving you a $4$-dimensional solution space. Choose $m_6, m_5,m_3$ and $m_2$ freely, and back substitute to get $m_1$ and $m_2$...

Or, perhaps better,why not use $m_1=2$ and $m_5=1$, then set the other $m$'s to $0$?

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The columns of $M$ are images of the standard basis vectors, which correspond to the polynomials $x^2$, $x$ and $1$, in that order. $(x^2)'=2x$, so the first column of $M$ is $(2,0)^T$. Continue in this way with $x$ and $1$ to find the second and third columns, respectively, of $M$.

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