anonymous
  • anonymous
Pls help. see the attachment for the question.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@JamesJ can u help me pls?
JamesJ
  • JamesJ
Please is such an underused work on this site.

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JamesJ
  • JamesJ
Let u=i+j+2k, v=3i+j-k and w=βi+(β-1)j-k 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.
JamesJ
  • JamesJ
*word
anonymous
  • anonymous
I am sorry I didn't get you.
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
I.e., the row reduction of 1 1 2 β β-1 -1 will be the same as the row reduction of 3 1 -1 β β-1 -1 Use that fact to find a value of β for which that is true.
anonymous
  • anonymous
If you dnt mind can u tel me vat is row deduction?
JamesJ
  • JamesJ
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
JamesJ
  • JamesJ
Now equate those two expressions for the vector x and find what what implies for beta.
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
Nw I understand it . Thanx a lot. was the way I askd u for the help wrong? If so I am extremely sorry.
JamesJ
  • JamesJ
You asked *correctly*, using the word 'please'. Please is a very good word.
anonymous
  • anonymous
Thanx a lot again.
anonymous
  • anonymous
hw do arbitary constants limit the choice of beta
anonymous
  • anonymous
|dw:1331146251985:dw| After this I am stuck. Can you please tell me how to proceed after this?
JamesJ
  • JamesJ
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?

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