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Prove that if $x^{k+1}=Bx^{k}+c$, converge to the solution of $Ax=b$ then $c=(Id-B)A^{-1}b$

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Taking limit on $x_{n+1}=Bx_n+c$ as $n\to \infty$ $$x=Bx+c\Rightarrow c=(I-B)x=(I-B)A^{-1}b.$$

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  • $\begingroup$ Where is the $d$ of $(Id-B)A^{-1}b$? $\endgroup$ – JuanMuñoz Aug 23 at 14:33
  • $\begingroup$ There's no $d$, like $d=1$. Anything in the solution unclear? Maybe it has a mistake somewhere? Thanks. $\endgroup$ – Alexey Burdin Aug 23 at 14:39

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