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Let A be an invertible n x n-matrix. The natrix A(1 : k, 1: k) is invertible for k = 1 Fation, we have det (A(1: k, 1 k)) = π₁₂. Tk, •πk, k= ot obtained after k-1 elimination steps. T

Let A be an invertible n x n-matrix. The natrix A(1 : k, 1: k) is invertible for k = 1 Fation, we have det (A(1: k, 1 k)) = π₁₂. Tk, •πk, k= ot obtained after k-1 elimination steps. T

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let A be an invertible n × n-matrix. Then A has an LU-factorization A = LU iff every matrix A(1 : k, 1: k) is invertible for k = 1,...,n. Furthermore, when A has an LU-factorization, we have det (A(1: k, 1 k)) = π₁₂. Tk, 1 k = 1,..., n, where T is the pivot obtained after k-1 elimination steps. Therefore, the kth pivot is given by a11 = if k = 1 Tk = det (A(1:1, 1:1)) det (A(1 : k, 1 k)) det(A(1 : k1, 1: k-1)) if k = 2,..., n.
Transcribed Image Text:Let A be an invertible n × n-matrix. Then A has an LU-factorization A = LU iff every matrix A(1 : k, 1: k) is invertible for k = 1,...,n. Furthermore, when A has an LU-factorization, we have det (A(1: k, 1 k)) = π₁₂. Tk, 1 k = 1,..., n, where T is the pivot obtained after k-1 elimination steps. Therefore, the kth pivot is given by a11 = if k = 1 Tk = det (A(1:1, 1:1)) det (A(1 : k, 1 k)) det(A(1 : k1, 1: k-1)) if k = 2,..., n.
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