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February 18th, 2017, 06:57 AM   #1
ZMD
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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
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February 18th, 2017, 08:00 AM   #2
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what have you tried?
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February 18th, 2017, 08:16 AM   #3
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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)
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February 18th, 2017, 09:53 PM   #4
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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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