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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 and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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1 -31 Forward elimination changes-1 -4 2 x = b to a triangular -5 2 [1 1 -37 0 -3 -1 x = C: 0 0 -3 1x + 1y3z=-6 - 1x - 4y + 2z = -5 ⇒ -2x - 5y + 2z = -5 Check that c = 9. The equation - 3z = -6 in Ux= c comes from the original 1x + 1y - 3z = -6 in Ax=b by subtacting 31 = times equation 1 and 32 = 9. times the final equation 2. Reverse that to recover 9. 1 in the last row of [A b] from the final [1 1 -3 -6] and [0 -3 -1 -11] and [0 03-6] 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? 9. 9. -2 8. 1x + 1y3z-6 -3y-12-11 → -3y - 4z = -17 1x + 1ly-3z=-6 -3y - 12 = -11 -32-6 9. C= X= -6 -11 solves the first one. Which x solves the second one? x = -6
Transcribed Image Text:1 -31 Forward elimination changes-1 -4 2 x = b to a triangular -5 2 [1 1 -37 0 -3 -1 x = C: 0 0 -3 1x + 1y3z=-6 - 1x - 4y + 2z = -5 ⇒ -2x - 5y + 2z = -5 Check that c = 9. The equation - 3z = -6 in Ux= c comes from the original 1x + 1y - 3z = -6 in Ax=b by subtacting 31 = times equation 1 and 32 = 9. times the final equation 2. Reverse that to recover 9. 1 in the last row of [A b] from the final [1 1 -3 -6] and [0 -3 -1 -11] and [0 03-6] 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? 9. 9. -2 8. 1x + 1y3z-6 -3y-12-11 → -3y - 4z = -17 1x + 1ly-3z=-6 -3y - 12 = -11 -32-6 9. C= X= -6 -11 solves the first one. Which x solves the second one? x = -6
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