Question

Show all Polygonal Sequences can be generated by solving the Matrix equation `Avec(x)= vec(b)` where `A` is `[[1, 1, 1], [4, 2, 1], [9,3,1]]` and `vec(b)=[[a_1], [a_2], [a_3]]` is the column vector? Show that `vec(x) =A^-1vec(b)` for all sequences?

Answer 1

See answer below for details:
`A^-1=[[1/2,-1,1/2],[-5/2,4,-3/2],[3,-3,1]]`
`vecb= [[1],[2+d],[3+3d]]`

and solution for `AA d in ZZ`

`[[a],[b],[c]]=[[d/2],[1-3/2d],[0]]`

Explanation:

Solution Strategy : Define arithmetic sequnce as
`a_n = {1, 1+d, 1+2d, 1+3d, cdots, 1+(n-1)d}`
a polygonal sequence is given nth sum of the arithmetic sequence:
`P_n^d = {1, 2+d, 3+3d, 4+6d, cdots)}`
Now let's choose d=1, Triangular Sequence:
`P_n^3 = {1, 3, 6, 10, cdots}`
It can be shown that : `P_n^3 = {1, 3, 6, 10, cdots}` can be generated by:
`P_n^3 = an^2+bn+c`
For n values of `n={1,2,3}` evaluate the polynomila an match with frist 3 entries of the Triangular Sequence:
`P_1^3 = a+b+c = 1`
`P_2^3 = a=4a+2b+c = 3`
`P_3^3 = 9a+3b+c = 6`
We have 3 equation and 3 uknown that can be rewritten in a vector matrix form:
`[[1,1,1],[4,2,1],[9,3,1]]xx[[a],[b],[c]]=[[1],[3],[6]]`
and in general:
`[[1,1,1],[4,2,1],[9,3,1]]xx[[a],[b],[c]]=[[1],[2+d],[3+3d]]`

This of the form `color(red)(A)vec(x)` and to solve for `vecx`
`vecx=color(red)(A)^-1vecb = color(red)(A)^-1 [[1],[2+d],[3+3d]]`

`[[a],[b],[c]]=[[1/2,-1,1/2],[-5/2,4,-3/2],[3,-3,1]]xx[[1],[2+d],[3+3d]]`
`[[a],[b],[c]]=[[d/2],[1-3/2d],[0]]`