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@JamesJ can u help me pls?
Please is such an underused work on this site.
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.
I am sorry I didn't get you.
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.
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.
If you dnt mind can u tel me vat is row deduction?
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
Now equate those two expressions for the vector x and find what what implies for beta.
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.
Nw I understand it . Thanx a lot. was the way I askd u for the help wrong? If so I am extremely sorry.
You asked *correctly*, using the word 'please'. Please is a very good word.
Thanx a lot again.
hw do arbitary constants limit the choice of beta
|dw:1331146251985:dw| After this I am stuck. Can you please tell me how to proceed after this?
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?