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Linear Algebra And Its Application*4th Edition

 
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MessagePosté le: Jeu 2 Juin - 01:24 (2016)    Sujet du message: Linear Algebra And Its Application*4th Edition Répondre en citant




Linear Algebra And Its Application 4th Edition > urlin.us/2jdip























































Linear Algebra And Its Application 4th Edition

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If A has m pivot positions, then m pivot columns, each containing one pivot position. Comment(0) Step 23 of 26 w) False The statement would be true if the condition is not zero were present. Step 1 of 26 (a) False (The word reduced is missing) Counter example The matrix A is row equivalent to matrices B and C, both is echelon form. Since B is row equivalent to A, B can be transformed by row operations first into A and then further transformed into U. Comment(0) Step 10 of 26 j) False Every equation has the trivial solution whether or not some variables are free. Comment(0) Step 25 of 26 y) True For the transformation to map onto, the matrix A would have to have a pivot in every row and hence have six pivot columns. Comment(0) Step 21 of 26 u) False If u and v are multiples, then Span is a line, and w need not be on that line.

Comment(0) Step 4 of 26 d) False Counter example: The following system has no free variables and no solution: Comment(0) Step 5 of 26 e) True If is transformed into by elementary row operations, then the two augmented matrices are row equivalent. Then the equation is consistent and has m basic variables and at least one free variable. Comment(0) Step 17 of 26 q) True If none of the vectors in the set in is a multiple of one of the other vectors, then S is linearly independent. Comment(0) Step 18 of 26 r) True Any set of three vectors in would have to be linearly dependent, by Theorem 8 in Section 1.6 Comment(0) Step 19 of 26 s) False If as set were to span, then the matrix would have a pivot position in each of its five rows, which is impossible since A has only four columns. By the Existence and Uniqueness Theorem in Section 1.2, the system has infinitely many solutions. If A is an m n matrix, if the equation Ax = b has at least two different solutions, and if the equation Ax = c is consistent, then the equation Ax = c has many solutions. Then , so w is a linear combination of u and v. g.

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