Math 104, Summer 2010 Homework 6 Solutions Note: we only
... Thus kA−1 k1 ≈ 2 · 1010 , kA−1 k∞ ≈ 1010 , and kA−1 kF ≈ 2 · 1010 . Thus, with respect to these three norms, the condition numbers κ(A) = kAkkA−1 k are approximately 2 · 1010 , 1010 , and 2 · 1010 . These are all large numbers, showing that the problem of solving Ax = b for this particular A is ill- ...
... Thus kA−1 k1 ≈ 2 · 1010 , kA−1 k∞ ≈ 1010 , and kA−1 kF ≈ 2 · 1010 . Thus, with respect to these three norms, the condition numbers κ(A) = kAkkA−1 k are approximately 2 · 1010 , 1010 , and 2 · 1010 . These are all large numbers, showing that the problem of solving Ax = b for this particular A is ill- ...
