homework_9_solutions - MA 265 HOMEWORK ASSIGNMENT#9...

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Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #9 SOLUTIONS #1. Page 282; Exercise 1. Find a basis for the subspace V of R 3 spanned by S =      1 2 3   ,   2 1 4   ,  - 1- 1 2   ,   1 2   ,   1 1 1      and write each of the following vectors in terms of the basis vectors: (a)   3 4 12   (b)   3 2 2   (c)   1 2 6   Solution: In order to compute a basis for V = span S , we form the matrix with the elements of S as its columns, then compute the reduced column echelon form:   1 2- 1 0 1 2 1- 1 1 1 3 4 2 2 1   = ⇒   1 0 0 0 0 0 1 0 0 0 0 0 1 0 0   . By considering the first three columns, we see that a basis for V is      1   ,   1   ,   1      (a) We may choose the following linear combination:   3 4 12   = 3   1   + 4   1   + 12   1   (b) We may choose the following linear combination:   3 2 2   = 3   1   + 2   1   + 2   1   (c) We may choose the following linear combination:   1 2 6   = 1   1   + 2   1   + 6   1   1 2 MA 265 HOMEWORK ASSIGNMENT #9 SOLUTIONS #2. Page 282; Exercise 3. Find a basis for the subspace M 22 spanned by S = 1 2 1 1 , 2 1 3 1 , 0 2 1 2 , 3 2 1 4 , 5- 1 . Solution #1: We determine which matrices can be expressed as linear combinations of the others. To this end, consider the equation a 1 1 2 1 1 + a 2 2 1 3 1 + a 3 0 2 1 2 + a 4 3 2 1 4 + a 5 5- 1 = 0 0 0 0 . We can express this as the linear system a 1 + 2 a 2 + 3 a 4 + 5 a 5 = 0 2 a 1 + a 2 + 2 a 3 + 2 a 4 = 0 a 1 + 3 a 2 + a 3 + a 4 = 0 a 1 + a 2 + 2 a 3 + 4 a 4- a 5 = 0 = ⇒     1 2 0 3 5 2 1 2 2 1 3 1 1 1 1 2 4- 1     . The reduced row echelon form for the augmented matrix is       1 0 0 0 13 5 0 1 0 0 0 0 1 0- 17 5 0 0 0 1 4 5       = ⇒ a 1 + 13 5 a 5 = 0 a 2 = 0 a 3- 17 5 a 5 = 0 a 4 + 4 5 a 5 = 0 Since a 5 can be chosen as an arbitrary variable, we see that the last matrix in S can be expressed as a linear combination of the others. In fact, there is only one free variable in this linear system, so a basis would be 1 2 1 1 , 2 1 3 1 , 0 2 1 2 , 3 2 1 4 Solution #2: The real vector space M 22 has dimension 4. Hence the subset above is also a basis for M 22 , so another choice of basis is the standard basis for M 22 : 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 #3. Page 282; Exercise 6. In Exercises 5 and 6, find a basis for the row space of A consisting of vectors that (a) are not necessarily row vectors of A ; and (b) are row vectors of A ....
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

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homework_9_solutions - MA 265 HOMEWORK ASSIGNMENT#9...

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