Chapter 3 Solution of Linear Systems - Math/CS

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66 CHAPTER 3. SOLUTION OF LINEAR SYSTEMS If we can compute the factorization P A = LU, then Ax = b ⇒ P A = P b ⇒ LUx = P b ⇒ Ly = P b, where Ux = y. Therefore, the following steps can be used to solve Ax = b: • Compute the factorization P A = LU. • Permute entries of b to obtain d = P b. • Solve Ly = d using forward substitution. • Solve Ux = y using backward substitution. Example 3.3.5. Use the P A = LU factorization of the previous example to solve Ax = b, where ⎡ ⎤ 1 b = ⎣ 2 ⎦ 3 Since the P A = LU factorization is given, we need only: ⎡ • Obtain d = P b = ⎣ 3 ⎤ 1 ⎦ 2 • Solve Ly = d, or ⎡ 1 0 ⎤ ⎡ 0 ⎣ 1 7 1 0 ⎦ ⎣ 1 Using forward substitution, we obtain y1 = 3 1 7y1 + y2 = 1 ⇒ y2 = 1 − 1 4 7 (3) = 7 4 7y1 + 1 2y2 + y3 = 2 ⇒ y3 = 2 − 4 1 7 (3) − 2 • Solve Ux = y, or ⎡ 7 8 9 ⎣ 0 0 6 7 0 19 7 − 1 2 Using backard substitution, we obtain − 1 2 x3 = 0 ⇒ x3 = 0. 6 7x2 + 19 7 x3 = 4 7 ⇒ x2 = 7 6 ( 4 7 4 7 7x1 + 8x2 + 9x3 = 3 ⇒ x1 = 1 7 1 2 y1 y2 y3 ⎤ 4 ( 7 ) = 0 ⎤ ⎡ ⎦ 19 2 − 7 (0)) = 3 . (3 − 8( 2 3 Therefore, the solution of Ax = b is given by ⎡ x = ⎣ ⎣ x1 x2 x3 ⎡ ⎦ = ⎣ ⎤ ⎡ ⎦ = ⎣ 3 1 2 ) − 9(0)) = − 1 3 . Problem 3.3.5. Use Gaussian elimination with partial pivoting to find the P A = LU factorization of the following matrices: − 1 3 2 3 0 ⎤ ⎦ . 3 4 7 0 ⎤ ⎦ ⎤ ⎦
3.4. THE ACCURACY OF COMPUTED SOLUTIONS 67 ⎡ 1 3 ⎤ −4 (a) A = ⎣ 0 −1 5 ⎦ 2 0 4 ⎡ (b) A = ⎣ 3 2 6 5 9 2 ⎤ ⎦ −3 −4 −11 ⎡ 2 ⎢ (b) A = ⎢ 0 ⎣ −1 − 0 3 3 1 4 1 1 2 1 4 − 1 2 − 2 5 2 1 ⎤ ⎥ ⎦ 13 ⎡ ⎢ Problem 3.3.6. Suppose A = ⎢ ⎣ 1 4 2 2 8 3 −3 12 2 ⎤ 4 −8 ⎥ 1 ⎦ and ⎡ ⎤ 3 ⎢ b = ⎢ 60 ⎥ ⎣ 1 ⎦ −3 −1 1 −4 5 . Suppose Gaussian elimination with partial pivoting was used to obtain the P A = LU factorization, where ⎡ 0 ⎢ P = ⎢ 0 ⎣ 1 1 0 0 0 0 0 ⎤ 0 1 ⎥ 0 ⎦ , ⎡ 1 ⎢ L = ⎢ −3/4 ⎣ 1/4 0 1 0 0 0 1 ⎤ 0 0 ⎥ 0 ⎦ , ⎡ 4 ⎢ U = ⎢ 0 ⎣ 0 8 5 0 12 10 −6 ⎤ −8 −10 ⎥ 6 ⎦ 0 0 1 0 1/2 −1/5 1/3 1 0 0 0 1 . Use this factorization (not the matrix A) to solve Ax = b. 3.4 The Accuracy of Computed Solutions Methods for determining the accuracy of the computed solution of a linear system are discussed in more advanced courses in numerical linear algebra. However, the following material qualitatively describes some of the factors affecting the accuracy. Consider a nonsingular linear system Ax = b of order n. The solution process involves two steps: GE is used to transform Ax = b into an upper triangular linear system Ux = c, and Ux = c is solved by backward substitution (say, row–oriented). We use the phrase “the computed solution” to refer to the solution obtained when these two steps are performed using floating–point arithmetic. Generally, the solution of the upper triangular linear system Ux = c is determined accurately, so we focus on comparing the solutions of Ax = b and Ux = c. Remember, the solutions of Ax = b and Ux = c are different because GE was performed using floating–point arithmetic. The key to comparing the solutions of Ax = b and Ux = c is to reverse the steps used in GE. Specifically, consider the sequence of elementary operations used to transform Ax = b into Ux = c. Using exact arithmetic, apply the inverse of each of these elementary operations, in the reverse order, to Ux = c. The result is a linear system A ′ x = b ′ and, because exact arithmetic was used, it has the same solution as Ux = c. One of the nice properties of GE is that, generally, A ′ is “close to” A and b ′ is “close to” b. This observation is described by saying that GE is generally stable. To summarize: GE is generally stable. When GE is stable, the computed solution of Ax = b is the exact solution of A ′ x = b ′ where A ′ is close to A and b ′ is close to b. The next step is to consider what happens to the solution of Ax = b when A and b are changed a little. Linear systems can be placed into two categories. One category, consisting of so-called well–conditioned linear systems, has the property that all small changes in the matrix A and the right hand side b lead to a small change in the solution x of the linear system Ax = b. If a linear system is not well-conditioned then it is said to be ill–conditioned. So, an ill–conditioned linear system has the property that some small changes in the matrix A and/or the right hand side b lead
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Chapter 3 Solution of Linear Systems - Math/CS

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