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February 18th, 2017, 06:57 AM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1  Stumped. How to do this?
Given an nxnmatrix A with entries aij = (10·i)^j1 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: 379 Thanks: 205 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: 73 Thanks: 1 
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: USA Posts: 1,944 Thanks: 1011 
since they give you $x$ it looks like you just have to do the matrix multiplication to obtain $b$ let $i,j =1,2,\dots,n$ $A_{i,j}=(10i)^{j1}$ $x_j = 1$ $b_i = (Ax)_i = \displaystyle{\sum_{j=1}^n}~(10i)^{j1}= \displaystyle{\sum_{j=0}^{n1}}~(10i)^j= \dfrac{(10i)^n1}{10i1}$ so for example if $n=4$ $b = (1111, 8421, 27931, 65641)$ 

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linear system equations, stumped 
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