Forward elimination changes-1 [1 0 0 1 -31 -3 -1 x = C: 0 -3 1x + 1y3z = -6 - 1x - 4y + 2z = -5 ⇒ -2x - 5y + 2z = -5 Check that c= 1 1. 9. 1 -31 -4 2 x=b to a triangular -2 -5 2 1x + 1y3z-6 The equation -32= -6 in Ux= c comes from the original 1x + 1y - 3z = -6 in Ax=b by subtacting 31 = O times equation 1 and 32 = 9. times the final equation 2. Reverse that to recover [ ] in the last row of [A b] from the final [1 1 -3-6] and [0-3 -1 -11] and [00-3-6] in [U c]. Row 3 of [A b] = (31 Row 1 + 32 Row 2 + 1 Row 3) of [U c]. In matrix notation this is multiplication by L. So A = LU and b = Lc. What are the 3 by 3 triangular systems Lc = b and Ux = c? 41 -3y-12-11 → -3y4z=-17 1x + ly 3z = -6 -3y1z=-11 -32=-6 9. 8. C= X= -6 8 -11 solves the first one. Which x solves the second one? x = -6
Forward elimination changes-1 [1 0 0 1 -31 -3 -1 x = C: 0 -3 1x + 1y3z = -6 - 1x - 4y + 2z = -5 ⇒ -2x - 5y + 2z = -5 Check that c= 1 1. 9. 1 -31 -4 2 x=b to a triangular -2 -5 2 1x + 1y3z-6 The equation -32= -6 in Ux= c comes from the original 1x + 1y - 3z = -6 in Ax=b by subtacting 31 = O times equation 1 and 32 = 9. times the final equation 2. Reverse that to recover [ ] in the last row of [A b] from the final [1 1 -3-6] and [0-3 -1 -11] and [00-3-6] in [U c]. Row 3 of [A b] = (31 Row 1 + 32 Row 2 + 1 Row 3) of [U c]. In matrix notation this is multiplication by L. So A = LU and b = Lc. What are the 3 by 3 triangular systems Lc = b and Ux = c? 41 -3y-12-11 → -3y4z=-17 1x + ly 3z = -6 -3y1z=-11 -32=-6 9. 8. C= X= -6 8 -11 solves the first one. Which x solves the second one? x = -6
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 28RE
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