5x1 + 6x2 + 11x3 + 7x4 6x5 5 4. Membership in a Span: For every item: (i) apply the Modified Gauss-Jordan Algorith to determine if b is in Span(S); (ii) write the solution vector i if there are solutions to ti corresponding augmented matrix; (iii) express b as a linear combination of the vectors S in the simplest possible way, and check directly that your answer is correct. b = (-4, 2,–3); S = {(7, 4,–6), (–5,–2, 3)} b. b = (-4, 2,–4); S = {(7,4,–6), (-5,–2, 3)} %3| %3D b = (9,7,–8, 2); S = {(5,-3, 2, 6), (-2,–3, 5, 8), (–5, 4,-2,–3)} d. b = (-10, 13,–4, 9); S = {(5,-3, 2, 6), (–5, 4,–2,–3), (–5, 7,–2, 6)} с. a.
5x1 + 6x2 + 11x3 + 7x4 6x5 5 4. Membership in a Span: For every item: (i) apply the Modified Gauss-Jordan Algorith to determine if b is in Span(S); (ii) write the solution vector i if there are solutions to ti corresponding augmented matrix; (iii) express b as a linear combination of the vectors S in the simplest possible way, and check directly that your answer is correct. b = (-4, 2,–3); S = {(7, 4,–6), (–5,–2, 3)} b. b = (-4, 2,–4); S = {(7,4,–6), (-5,–2, 3)} %3| %3D b = (9,7,–8, 2); S = {(5,-3, 2, 6), (-2,–3, 5, 8), (–5, 4,-2,–3)} d. b = (-10, 13,–4, 9); S = {(5,-3, 2, 6), (–5, 4,–2,–3), (–5, 7,–2, 6)} с. a.
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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Transcribed Image Text:5x1 + 6x2 + 11x3 +
7x4
6x5
5
4. Membership in a Span: For every item: (i) apply the Modified Gauss-Jordan Algorith
to determine if b is in Span(S); (ii) write the solution vector i if there are solutions to ti
corresponding augmented matrix; (iii) express
S in the simplest possible way, and check directly that your answer is correct.
as a linear combination of the vectors
а.
Б - (-4, 2, -3); S - {{7,4, -6), (-5, -2, 3)}
b. b = (-4, 2,–4); S = {(7,4,–6), (-5,–2, 3)}
%3|
b = (9, 7,-8, 2); S = {(5,-3, 2, 6), (-2,–3, 5, 8), (–5, 4,–2,–3)}
d. b = (-10, 13,–4, 9); S = {(5,-3, 2, 6), (-5, 4,–2,–3), (-5, 7,–2, 6)}
Б - (13, 14,-18,-11);
S = {(5,-3, 2, 6), (–2,–3, 5, 8), (–5, 4,–2,–3), (11, 10,–13,-6)}
b = (8,–9,–8, 15,-3), S = {(6,0,4, 3, 2), (3, 2, 7, 1,–2), (2, 1, 2, –1, 3)}
b = (-4,–1, 4, 7,-9), S = {(3, 2, 7, 1,–2), (2, 1, 2,–1, 1), (1, 0,–3,–3, 4)}
h. Б - (-3,6, -1,-9, 4),
S = {(6,0,–1, 3, 2), (3, 2,–3, 1,–2), (0,–4, 5, 1, 6), (3,–2, 1, 3, 3)}
с.
%3D
е.
f.
g.
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