Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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- Please solve for P* Q* and v 11.3 3arrow_forwardIs it possible that all solutions of a homogeneous system of twelve linear equations in fourteen variables are multiples of one fixed nonzero solution? Discuss. Consider the system as Ax = 0, where A is a 12x 14 matrix. Choose the correct answer below. O A. Yes. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, dim Nul A = 14- rank Az 2. Since there is at least one free variable in the system, all solutions are multiples of one fixed nonzero solution. O B. No. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A = 12 - rank A= 0. Since Nul A = 0, it is impossible to find a single vector in Nul A that spans Nul A. OC. No. Since A has at most 12 pivot positions, rank As 12. By the Rank Theorem, Nul A= 14 - rank Az 2. Thus, it is impossible to find a single vector in Nul A that spans Nul A. O D. Yes. Since A has 12 pivot positions, rank A= 12. By the Rank Theorem, dim Nul A= 12- rank A= 0. Thus, all solutions are multiples of one fixed nonzero…arrow_forwardSolve the following systems of equations using a valid elimination algorithm, include all the elementary line operations that you used to solve it, then check that your result is correct, using another method (it can be computational). Write the solution vector. note: Do everything that is asked above (in the image are the equations and the matrix)arrow_forward
- Any two row equivalent matrices have the same reduced row echelon forms. A True B) Falsearrow_forward5x1 + 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.arrow_forward
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