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atjari Group TitleBest ResponseYou've already chosen the best response.0
@JamesJ can u help me pls?
 2 years ago

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Let 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.
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
I am sorry I didn't get you.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
I 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.
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
I.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.
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
If you dnt mind can u tel me vat is row deduction?
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Ok, 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
 2 years ago

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
You 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.
 2 years ago

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

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

atjari Group TitleBest ResponseYou've already chosen the best response.0
Thanx a lot again.
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
hw do arbitary constants limit the choice of beta
 2 years ago

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.1
Think 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?
 2 years ago
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