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anonymous
 4 years ago
Pls help. see the attachment for the question.
anonymous
 4 years ago
Pls help. see the attachment for the question.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@JamesJ can u help me pls?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Please is such an underused work on this site.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Let u=i+j+2k, v=3i+jk and w=βi+(β1)jk where β is a real number. Find β such that the space spanned by u and v is the same as the space spanned by u and w.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am sorry I didn't get you.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1I am saying thank you for asking with "please", vs. just putting out a demand that I or someone else help you. It's slightly ironic. Don't worry about it. Down to business. If u and v span the same space as u and w, then they will have the same row reduction.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1I.e., the row reduction of 1 1 2 β β1 1 will be the same as the row reduction of 3 1 1 β β1 1 Use that fact to find a value of β for which that is true.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If you dnt mind can u tel me vat is row deduction?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Ok, you don't that. Another strategy then. If the span of the two pairs of vectors is the same, then any vector in one span can be written as the vector in the other. Hence for any vector in the span of u and w, i.e., x = au + bw where a and b are arbitrary constants. x is also a member of the span of v and w; i.e., there exist constants c and d such that x = cv + dw

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Now equate those two expressions for the vector x and find what what implies for beta.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1You will end up with three equations, one for each component of x. Remember that a and b are arbitrary. This constrains your choice of beta.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nw I understand it . Thanx a lot. was the way I askd u for the help wrong? If so I am extremely sorry.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1You asked *correctly*, using the word 'please'. Please is a very good word.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hw do arbitary constants limit the choice of beta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1331146251985:dw After this I am stuck. Can you please tell me how to proceed after this?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Think of this now as a system of equations in three variables: c, d and beta. a and b are just parameters. Now ask yourself the question: what is the solution to this system of equations in those three variables?
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