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Suppose there is a matrix equation such that:

$Ax = b$

where $A$ is given, which is an upper triangular matrix, and $b$ is given.

Then, $x$ and $b$ are perturbed by vectors $p$ and $q$ such that:

$A(x+p) = (b+q)$

The question asks to solve for $p$ in terms of $q$.

To approach this, I am thinking of finding $x$ by using the first equation, and then plug that $x$ into the second equation. Then, I can solve $p$ in terms of $q$. Do you think this (plugging in $x$ to the second equation) is acceptable? That is, is $x$ the same in both equations?

1 Answers 1

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Matrix multiplication distributes over addition... So:

A(x+p) = b+q Ax + Ap = b + q 

But since Ax = b, subtract that from the last equation to get:

Ap = q 

If A is upper-triangular, solving this for p in terms of q is simple using back substitution.

(Am I missing something?)

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    +! - a terrific answer.2011-06-06