Suppose that the linear system $A\mathbf{x}=\mathbf{b}$ is perturbed so that $(A+\delta A)\mathbf{x}=\mathbf{b}$.
We can calculate the relative error $\frac{\|\mathbf{x-\bar{x}\|}}{\|\mathbf{x}\|}$ if we know the true value of $\mathbf{x}$ and have an estimate for $\mathbf{\bar{x}}$, but if we want to put a general bound on the relative error in the system in terms of the system's condition number $K(A) = \|A^{-1}\|\cdot\|A\|$ and $\|\delta A\|/\|A\|$, how can this be done?
I have been trying to rearrange what is known in order to gain an expression in terms of this, and have so far obtained, with working:
$A\mathbf{x}=(A+\delta A)\mathbf{\bar{x}}$ equating the given formulae
$(A+\delta A)^{-1}A\mathbf{x}=\mathbf{\bar{x}}$ multiplying both sides by inverse
$(A+\delta A)^{-1}(A^{-1})^{-1}\mathbf{x}=\mathbf{\bar{x}}$ rewriting $A = (A^{-1})^{-1}$
$((A^{-1})(A+\delta A))^{-1}\mathbf{x}=\mathbf{\bar{x}}$ rearranging the inversion
$(I + A^{-1}\delta A)^{-1}\mathbf{x}=\mathbf{\bar{x}}$ multiplying out the brackets
At this point I become a bit dubious about the next step. I know that at some point the left hand side will need to look like ${\|\mathbf{x-\bar{x}\|}}$ so as to rearrange for the relative error, but multiplying the last line above by $-1$ then adding $\mathbf{x}$ to both sides, then taking the norm to obtain:
$\|\mathbf{x}-(I + A^{-1}\delta A)^{-1}\mathbf{x}\|=\|\mathbf{\mathbf{x}-\bar{x}}\|$
seems a bit too convenient and following this through with pen and paper doesn't seem to lead anywhere. Is this the correct approach, and if not, is there a chance of some guidance towards the correct manner?