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# sol2 - cep 2008/09 MT3802 Numerical Analysis SOLUTIONS...

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cep 2008/09 MT3802 - Numerical Analysis SOLUTIONS - Tutorial Sheet 2 1. For a sub-ordinate matrix norm and an invertible matrix A , where Ax = b , show that ( i ) A s A s s N Using Rule 5 (1.9) of sub-ordinate matrix norms we have A s A A s - 1 A 2 A s - 2 ≤ · · · ≤ A s - 1 A A s ( ii ) 1 A x b A - 1 Consider the linear system Ax = b Ax = b A x 1 A x b (1) and similarly, Ax = b x = A - 1 b x = A - 1 b A - 1 b (2) Combining (1) & (2) we get: 1 A x b A - 1 . 2. Assume that ( A + δ A ) - 1 is computed as an approximation to A - 1 . Show that ( A + δ A ) - 1 - A - 1 = - ( A + δ A ) - 1 A . A - 1 = - ( I + A - 1 δ A ) - 1 . A - 1 δ A . A - 1 . (3)

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To find the first expression we can write the LHS as ( A + δ A ) - 1 - A - 1 = ( A + δ A ) - 1 [ I - ( A + δ A ) A - 1 ] = ( A + δ A ) - 1 [ I - I - δ A A - 1 ] . = - ( A + δ A ) - 1 δ A A - 1 The second expression is gained by rewriting the RHS as ( A + δ A ) - 1 - A - 1 = - ( A + δ A ) - 1 δ A A - 1 = - [ A ( I + A - 1 δ A )] - 1 δ A A - 1 = - ( I + A - 1 δ A ) - 1 A - 1 δ A A - 1 Provided that γK ( A ) < 1, show that ( A + δ A ) - 1 - A - 1 A - 1 γK ( A ) 1 - γK ( A ) (4) where γ = δ A / A and K ( A ) is the condition number of the matrix A . Hint: Use the identity from the first part of the question and remember (from Section 1.7.1) that
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