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I don't really know what counts as a proof and haven't been taught maths since 16yo (29yo PhD now). I've got working knowledge of e.g. basic linear algebra, geometry, and statistics, but this feels like running before I can walk so I'm now going back and filling in the elementary gaps. I'm working through a linear algebra textbook for my own learning.

I have no idea what one of the first exercises is asking for (q7): linear algebra textbook exercise

(Ignoring that the book says linear equations don't contain powers greater than 1, yet here is a 'system of linear equations' for a quadratic function...)

My idea for 'showing' this entails flipping to the previous page and pointing to the bit where they introduce augmented matrices as another notation for systems of linear equations. That juxtaposition of $x^{n}_{i}$, $a$, $b$, $c$ and $y_{i}$ equates to that augmented matrix because you told me so, therefore it is trivially and arbitrarily the case. Like how I've rote-learned Hund is dog in German and would 'show' this using a German-English dictionary.

So instead I tried solving using the steps they described: multiplying rows by constants and/or adding rows to each other. But no $x$ or $y$ appears more than once per row, so nothing can be eliminated, canceled or otherwise simplified this way.

I get the 'feel' for a lot of this, e.g. the utility of augmented matrices as a compact representation of systems of equations, and how they relate to the geometric interpretation of many points on one curve (even if it isn't linear). I just don't get the thrust of the question. Math.SE says 'show that' means 'prove that', but this proof seems to require zero deductive steps after the initial definition.

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    $\begingroup$ It means you have to solve for the system of equations $$\begin{cases}ax_1^2+bx_1+c=y_1\\ax_2^2+bx_2+c=y_2\\ax_3^2+bx_3+c=y_3\end{cases}$$ in which the unknowns are $a,b,c$. It indeed is linear in $a,b,c$. $\endgroup$ Commented Dec 8, 2015 at 0:28
  • $\begingroup$ Suppose you were just given the points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$. How would you normally find the values of $a,b,c$? $\endgroup$ Commented Dec 8, 2015 at 0:28
  • $\begingroup$ In theory, Gauß's pivot method. However, using linear algebra, there are more clever methods. $\endgroup$ Commented Dec 8, 2015 at 0:31

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The equation of the parabola is $y = ax^2 + bx + c$. Given that we know three points $(x_n,y_n)$, we can write each one out like so:

\begin{matrix} \\ ax_1^2 + bx_1 + c = y_1 \\ ax_2^2 + bx_2 + c = y_2 \\ ax_3^2 + bx_3 + c = y_3 \end{matrix}

But hey, we can also represent that in matrix form like so:

\begin{bmatrix} x_1^2 & x_1 & 1 &| y_1\\ x_2^2 & x_2 & 1 &|y_2 \\ x_3^2 & x_3 & 1 &|y_3 \end{bmatrix}

Where multiplying by the constants as a vector gives us back the system. In other words, we're solving for $Ax = x'$ which is done by solving the augmented matrix.

This might be easier to see if we try an example. Imagine we have three points $(0, -1), (2, 7)(-1, -2)$ Solving the matrix:

\begin{bmatrix} 0 & 0 & 1 &| -1\\ 4 & 2 & 1 &|7 \\ 1 & -1 & 1 &|-2 \end{bmatrix}

Will give back the constants, allowing us to write the equation.

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The "show that" process is fairly trivial. It amounts to turning the constraints "passes through" into equations involving the unknowns $a,b,c$.

As we know that the equation of the curve is $y=ax^2+bx+c$, then the three constraints are written as $y_k=ax_k^2+bx_k+c$ and they form a system of $3$ equations in $3$ unknowns, which you can represent by an augmented matrix, in order to solve it in this standard form.

The proof essentially consists in recalling that a curve of a given equation goes through a point iff the coordinates of this point satisfy the equation.

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  • $\begingroup$ Just want to say thanks for revisiting my old question :) My difficulty back then was much more I-want-to-say epistemological, in the vein of this: math.stackexchange.com/q/397972/229219 $\endgroup$ Commented Oct 3, 2024 at 5:43
  • $\begingroup$ @benxyzzy: this can't be inferred from the question. :-( $\endgroup$ Commented Oct 3, 2024 at 7:40

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