Chapter 3 Solution of Linear Systems - Math/CS

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58 CHAPTER 3. SOLUTION OF LINEAR SYSTEMS ! Gaussian elimination – row oriented for k = 1 to n − 1 !. . . Exchange step !. . . Find p such that |ap,k| is the first of the !. . . largest values among {|aq,k|} n q=k p := k for i = k + 1 to n if |ai,k| > |ap,k| then p := i next i ! exchange equations p and k if necessary if p > k then for j = k to n temp := ak,j; ak,j := ap,j; ap,j := temp next j temp := bk; bk := bp; bp := temp endif if ak,k == 0 then return (linear system is singular) !. . . elimination step - row oriented for i = k + 1 to n mi,k := ai,k/ak,k !. . . compute multiplier ai,k := 0 !. . . start of elimination for j = k + 1 to n ai,j := ai,j − mi,k ∗ ak,j next j bi := bi − mi,k ∗ bk next i next k if an,n == 0 then return (linear system is singular) ! Backward substitution – row oriented for i = n downto 1 do for j = i + 1 to n bi := bi − ai,j ∗ xj next j xi := bi/ai,i next i Figure 3.9: Pseudocode row oriented GE and Backward Substitution
3.2. GAUSSIAN ELIMINATION 59 Example 3.2.2. Using exact arithmetic, 2 stages of GE transforms the following singular linear system of order 3: 1x1 + 1x2 + 1x3 = 1 1x1 + (−1)x2 + 2x3 = 2 3x1 + 1x2 + 4x3 = 4 into the triangular form: 3x1 + 1x2 + 4x3 = 4 0x1 + (−4/3)x2 + 2/3x3 = 2/3 0x1 + 0x2 + 0x3 = 0 which has an infinite number of solutions. When floating–point arithmetic is used, the zeros in the last equation will be very small numbers due to rounding, and the resulting computed upper triangular matrix will usually be nonsingular. Particularly, the pivot value −4/3 cannot be represented exactly in IEEE standard arithmetic, so operations involving it, like the computation of the final version of the third equation, are subject to rounding errors. Still, backward substitution would give a good approximation to one (essentially arbitrary choice of) solution of the original linear system. If the original first equation x1 + x2 + x3 = 1 was replaced by x1 + x2 + x3 = 2, then with exact arithmetic 2 stages of GE would yield 0x3 = 1 as the last equation and there would be no solution. In floating–point arithmetic, a suspiciously large solution may be computed because in this resulting last equation the coefficient 0 multiplying x3 would usually be computed as a non-zero, but very small, pivot and the right hand side 1 would be changed only slightly. Problem 3.2.6. In Fig. 3.9, in the for-j loop (the elimination loop) the code could be logically simplified by recognizing that the effect of ai,k = 0 could be achieved by removing this statement and extending the for-j loop so it starts at j = k instead of j = k + 1. Why is this NOT a good idea? Hint: Think of the effects of floating–point arithmetic. Problem 3.2.7. In Fig. 3.9, the elimination step is row–oriented. Change the elimination step so it is column–oriented. Hint: In effect, you must exchange the order of the for-i and for-j loops. Problem 3.2.8. In Fig. 3.9, show that the elimination step of the k th stage of GE requires n − k divisions to compute the multipliers, and (n − k) 2 + n − k multiplications and (n − k) 2 + n − k subtractions to perform the eliminations. Conclude that the elimination steps of GE requires about n−1 k=1 (n − k) = n−1 n(n − 1) k=1 k = 2 n−1 k=1 (n − k)2 = n−1 k=1 k2 n(n − 1)(2n − 1) = 6 n−1 k=1 (n − k)2 = n−1 k=1 k2 n(n − 1)(2n − 1) = 6 We say f(n) ∼ g(n) when the ratio f(n) g(n) ∼ n2 2 divisions approaches 1 as n becomes large. ∼ n3 3 multiplications ∼ n3 3 subtractions Problem 3.2.9. How many comparisons does GE use to find pivot elements? Problem 3.2.10. How many assignments would you use to accomplish all the interchanges in the GE algorithm? Hint: Assume that at each stage an interchange is required. 3.2.3 The Role of Interchanges When floating–point, rather than exact, arithmetic is used in Gaussian elimination, it is important to carefully choose each pivot. To illustrate this fact, apply Gaussian elimination to the linear system 0.000025x1 + 1x2 = 1 1x1 + 1x2 = 2 for which the exact solution is x1 = 40000 39999 ≈ 1 and x2 = 39998 39999 ≈ 1.
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