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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 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: 635 Thanks: 401 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: 2,531 Thanks: 1390 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 Tags linear system equations, stumped Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tyrean Algebra 4 August 20th, 2013 06:25 PM tangosix Algebra 2 March 2nd, 2010 10:39 AM yabo2k Advanced Statistics 1 October 27th, 2009 05:41 AM bretttolbert Algebra 4 April 5th, 2008 12:15 PM malikah Algebra 5 September 11th, 2007 08:38 AM

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