Let x 6 3 2 tutorial 4 week 5 solutions 2 3 1 4 6 0

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Unformatted text preview: athematics 3. Let X = 6 3 2 Tutorial 4 (week 5) — Solutions 2 3 1 , 4 −6 0 , Page 2 . Find the dimension of Span(X ). Solution Note that the subset containing the ﬁrst two vectors of X is linearly independent (since the vectors are not scalar multiples of each other), and so the dimension of Span(X ) is at least 2. Hence, the dimension is either 2 or 3, depending on whether or not X is linearly independent. Suppose that 4 6 2 0 a 3 + b 3 + c −6 = 0 , 2 1 0 0 for some a, b, c ∈ R. To ﬁnd all possible a, b, c we have to solve the system of linear equations corresponding to the following augmented matrix: 62 40 3 3 −6 0 21 00 10 20 0 1 −4 0 00 00 Row reduce −−− −−→ . (Note that the row echelon form of a matrix is not unique, as it depends upon the sequence of row operations that you choose; the reduced row echelon form is, however, unique. Consequently, you might obtain a different row echelon form for the augmented matrix above. However, your row echelon form should still give the same set of solutions as the one above. This comment applies to all of the questions below which compute row echelon forms.) Looking at the row reduced form of the augmented matrix above, we can take c to be a free variable. The general solutions is c = r, for any r ∈ R, b = 4c = 4r, and a = −2c = −2r. In particular, taking r = 1 we have that 6 3 2 −2 Hence X is linearly dependent and 2 3 1 2 3 1 +4 4 −6 0 + 4 −6 0 = 0 0 0 . can be written as a linear combination of 6 3 2 and . 6 3 2 So Span(X ) = Span of Span(X ) is 2. 4. Let X = 1 0 0 0 4 , 2 1 0 0 , 3 2 1 0 2 3 1 , , 4 −6 0 = Span 6 3 2 , 2 3 1 and the dimension . Show that X is a linearly independent subset of R4 . Does Span(X ) = R ? What is dim Span(X )? Solution Suppose a 1 0 0 0 +b 2 1 0 0 +c 3 2 1 0 = 0 0 0 0 , for some a, b, c ∈ R. Then a + 2b + 3c = 0 b + 2c = 0 c =0 There is only one solution of this linear system,...
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This note was uploaded on 06/04/2013 for the course MATH 1002 taught by Professor Cartwright during the One '08 term at University of Sydney.

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