Math 450 Midterm 3 1. Explain why there are numerical problems with solving the linear system 1 M M 1 1 2 by naive Gaussian elimination when M is large. Solution. Naive Gaussian forward elimination gives 1 M M 1 1 2 1 M M ∼ 0 1−M 2−M Then back substitution gives x2 = 2−M . For large enough M , this is x2 = 1 on the 1−M computer. In this case, the next step of back substitution gives x1 = 0. However, as M → ∞ the exact solution approaches x1 = x2 = 1. 1 2. Let A be a matrix of the form a11 a12 a13 a21 a22 a23 A= a31 0 a33 a41 0 0 a51 0 0 Write an ecient pseudocode for naive Gaussian Solution. a14 a24 a34 a44 0 a15 a25 a35 . a45 a55 forward elimination on A. An ecient pseudocode is for k = 1 : 4 for i = k + 1 : 5 c = aik /akk aik = 0 j =k+1:5 aij = aij − cakj end for end end (though it can be improved slightly to avoid six subtractions.) 2 3. Let A be an n × n (strictly) diagonally dominant matrix. (i) Must A have an LU factorization? Why or why not? Yes. Diagonal dominance is preserved during naive forward elimination on A. Thus, the diagonal entries must be nonzero at each step. This means naive forward elimination is successful, or equivalently, A has an LU factorization. (ii) Must A have a Cholesky factorization? Why or why not? Solution. No: A is not necessarily symmetric. 3 4. Find the LU factorization of the matrix Solution. 1 2 0 A = 1 1 1 2 0 1 . The factorization is 1 0 0 L = 1 1 0 , 2 4 1 1 2 0 U = 0 −1 1 . 0 0 −3 4 5. Carry out two iterations of the Jacobi method for Ax = b where A= using the initial guess x Solution. (0) 5 1 , 2 6 3 b= , 4 . What is the exact solution? 0 = 1 The iterations are x The exact solution is x = (1) = 2/5 , 2/3 x . 1/2 1/2 5 (2) = 7/15 . 8/15

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# Math 450 Midterm 3