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 February 18th, 2017, 06:57 AM #1 Member   Joined: Nov 2016 From: Kansas Posts: 48 Thanks: 0 Stumped. How to do this? Given an nxn-matrix A with entries aij = (10·i)^j-1 for i,j=1,....,n.Compute a vector b such that x =(1, 1,. .., 1) is the solution to the linear equation system Ax = b
 February 18th, 2017, 08:00 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 114 Thanks: 45 Math Focus: Dynamical systems, analytic function theory, numerics what have you tried?
 February 18th, 2017, 08:16 AM #3 Member   Joined: Nov 2016 From: Kansas Posts: 48 Thanks: 0 I came up with matrix A such that A= (1......... 10^n 1..........20^n . . 1..........(10n)^n) With such a matrix the only solution comes out to be the same as matrix x=(1,1,.....,1)
 February 18th, 2017, 09:53 PM #4 Senior Member     Joined: Sep 2015 From: CA Posts: 1,303 Thanks: 667 since they give you $x$ it looks like you just have to do the matrix multiplication to obtain $b$ let $i,j =1,2,\dots,n$ $A_{i,j}=(10i)^{j-1}$ $x_j = 1$ $b_i = (Ax)_i = \displaystyle{\sum_{j=1}^n}~(10i)^{j-1}= \displaystyle{\sum_{j=0}^{n-1}}~(10i)^j= \dfrac{(10i)^n-1}{10i-1}$ so for example if $n=4$ $b = (1111, 8421, 27931, 65641)$ Thanks from topsquark, Country Boy and ZMD

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