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[1 3 2 3 6 9 6 3 25. 1 2 4 1 2 2 4 9 1 2

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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Find the bases for the row space and column space of the matrix -2 5 0 3 5 -7 0 -6 3 -2 1 -3 -3 8 -9 1-9 A = 1 -2 consisting entirely of row vectors, or column vectors, from A. What are the rank and nullity of the matrix? Hint: You can use the RREF of A to find a basis of the column space. However, to find a basis of the row space of A consisting entirely of row vectors of A, you need to use the RREF of AT. For instance, let -(1) 2 2 2 B =
Prove that if the columns of A are linearly independent, then they must form a basis for col(A).
Suppose that A = [c, c2 C3] is a 3 × 3 matrix with columns c;, and suppose that a basis for the kernel 1 of A is given by the vector (a) Write and circle an equation involving the c; that shows that the c; are not linearly independent. (b) Since the kernel of A has dimension 1, the image of A has dimension 2, and therefore has a basis consisting of 2 of the c;. Which two make a basis of the image of A?
2. Evaluate: ONOO OONN 02 20 2 10 0 0 2 002 2 0 Do columns of this matrix form a basis of R4?
Solve this matrix problem, check the screenshot;
3. Find an example of a matrix that has Why can't this be a basis for the row space and the mullspace? 1. Explain why the vector v = H and as a basis for its row space and its column space. cannot be a row of A and also in the nullspace of A.
Find a basis for Col A and Nul A for given matrix A. A row echelon form is also provided: 1 2 -5 0 –-1 1 -5 -1 -8 4 3 1 4 -3 -9 -7 -2 1 | 3 10 -7 11 7 0 0 0
R is the reduced row echelon form of Q. Find a basis {b₁,b2, ..., bn} for the null space of Q. b₁ = b₂ Fill in vectors in numerical order. Leave blank those that are not needed. = b3 b4 = b5 = 6565 7070 -704 -306 4480 -1573 1694 168 74 - 1073 1579 2314 2492 -248 -108 8138 8764 872 -380 5553 -2613 3317 - 794 1169; R = 4111 -2814 280 122 -1783 - 1320 = 1 0 13 00 1 00 0 0 0 0 0 0 0 2 13 1 0 0 0 8 13 8 0 0 0 10 | 5 13 8 0 0
Determine the dimension of, and a basis for, the column space 2, 0 -4 -5 3 3 of the matrix A = 6. 3 1 1 3.