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Let's say I have the following equation $Ax=b$
My question is - how can i find a vector $b$ around which the above equation is not stable?
I have $$A= \begin{bmatrix} 1 & 0.999 \\ 1 & 1.001 \\ \end{bmatrix} $$

This is what i have done so far. $x=A^{-1}b$ and $$A^{-1}= \begin{bmatrix} 500.5 & -499.5 \\ -500 & 500 \\ \end{bmatrix} $$ So I see that the $\|A^{-1}\|=1000$ is very large. I am guessing that $\|b\|$ should be very small, but I do not know how to EXACTLY FORMALLY formulate the solution to this problem.

Thanks

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What do you mean by 'not stable'? –  Dennis Jaheruddin Jan 14 '13 at 12:22
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Forgive me if I misunderstand the problem, but I think you are looking for $b$ such that a small change in $b$ causes a big change in $x$. $b=(1,1)$ is promising; the solution $x$ is very close to $(0,0)$, but if you change $b$ a little to, say, $(2,1)$, then $x$ is quite far from $(0,0)$.

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