The curve (y = ax^2 + bx + c) shown in the accompanying figure passes through the points (x1,y1), (x2,y2) and (x3,y3) Show that the coefficients a, b, and c form a solution of the system of linear equations whose augmented matrix is
$$\begin{bmatrix}x_1^2 & x_1 & 1 & y_1 \\ x_2^2 & x_2 & 1 & y_2 \\ x_3^2 & x_3 & 1 & y_3 \end{bmatrix}$$
I was working the problem above from Elementary Linear Algebra: Applications Version. The way i solve it is by creating a quadratic equation that passes (x1, y1), (x2, y2), (x3, y3) and plug and chug to the augmented matrix to see whether or not it produces the exact equation. I got some question as to why this works, and does it work for any other type of equation?
First, an assumption on the visualization of the a linear system solution. Suppose a linear system with 2 equation and 2 unknowns. If it has single solution, the solution can be imagined as a single horizontal and vertical line that intersect in a 2d space. I assume what were actually doing with the augmented matrix is created a plane on 3d space and find the meeting point or solution for 3 of those plane, is this correct? If so this leads to the second question.
Second, if were creating a plane on the 3d space why does the 3rd column of the augmented matrix is (1, 1, 1) and not (0, 0, 0), which--i think--also implies, why does the linear equation for searching the coefficient must be in 3d space (ax + by + cz = p) and not just 2d line (ax + by = p)? why must the 3rd column of quadratic equation (ax^2 + bx + c = p) be included?
so in summary, we sort of put the quadratic equation in 3d space, created 3 planes that passes each (x1, y1), (x2, y2), (x3, y3), finding the solution of quadratic coefficent by searching for those planes intersection. Is this correct? If this can be done with any equation, why?
Im sorry if the question is hard to understand, or im not clear enough in my explanation or question.