The previous post I made was not well explained. I hope this new image would pass the information required across. So I'm trying to solve for the a's in terms of the other variables. The softwares I have available are; Matlab,Maple,Scientific work place.
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OdunayoO
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What mathematical software do you have at your disposal? – Bobson Dugnutt Jun 05 '17 at 15:23
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I pasted your matrix in, but it is still a poor question. You don't say what software you have available. Any software that does this symbolically has a nice manual you can look at. This is a routine invert the $7 \times 7$ matrix and multiply on the left problem. – Ross Millikan Jun 05 '17 at 15:27
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Do you really need software to do this ? By subtracting the first row from the second and the third form the ones below, you are halfway of an echelon form and only elimination of a $3\times3$ matrix remains. This is quite doable by hand. – Jun 05 '17 at 15:31
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@YvesDaoust can do by hand but too much computational errors. Using software makes it faster to compute – OdunayoO Jun 05 '17 at 15:42
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@pearlly_o: this is not true. A symbolic solver will give you general expressions, which will be huge, and quite error prone if you have to transliterate them. On the opposite, exploiting the specific structure of the matrix and working the Gaussian elimination, you obtain a very compact solution. – Jun 05 '17 at 15:51
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In the four last equations, move the constant coefficients $2,6,12,20$ to the unknowns to see a Vandermonde system appear. Solving it is equivalent to Lagrangian interpolation on four points, and it is advantageous to compute it by Neville's algorithm.
Then moving the unknowns $a_2$ to $a_5$ to the RHS, you end-up with a $2\times2$ system.
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By elimination between the last four equations, you get the system
$$\begin{bmatrix}x_1-x_0&x_1^2-x_0^2&x_1^3-x_0^3\\x_2-x_0&x_2^2-x_0^2&x_2^3-x_0^3\\x_3-x_0&x_3^2-x_0^2&x_3^3-x_0^3\end{bmatrix}\begin{bmatrix}6a_3\\12a_4\\20a_5\end{bmatrix}=\begin{bmatrix}f_1-f_0\\f_2-f_0\\f_3-f_0\end{bmatrix}$$ which you can readily solve. You can even factor as
$$\begin{bmatrix}1&x_1+x_0&x_1^2+x_1x_0+x_0^2\\1&x_2+x_0&x_2^2+x_2x_0+x_0^2\\1&x_3+x_0&x_3^2+x_3x_0+x_0^2\end{bmatrix}\begin{bmatrix}6a_3\\12a_4\\20a_5\end{bmatrix}=\frac1{(x_1-x_0)(x_2-x_0)(x_3-x_0)}\begin{bmatrix}f_1-f_0\\f_2-f_0\\f_3-f_0\end{bmatrix}$$
and by a single elimination step reduce to a $2\times2$ system
$$\begin{bmatrix}1&x_2+x_1+x_0\\1&x_3+x_1+x_0\end{bmatrix}\begin{bmatrix}12a_4\\20a_5\end{bmatrix}=\frac1{(x_1-x_0)(x_2-x_0)(x_3-x_0)(x_2-x_1)(x_3-x_1)}\begin{bmatrix}f_2-f_1\\f_3-f_1\end{bmatrix},$$ or
$$20a_5=\frac{f_3-f_2}{(x_1-x_0)(x_2-x_0)(x_3-x_0)(x_2-x_1)(x_3-x_1)(x_3-x_2)}.$$
The third equation yields
$$2a_2=f_0-6a_3x_0-12a_4x_0^2-20a_5x_0^3.$$
Then the first two form another easy $2\times2$ system.