IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: sahrhp on May 05, 2010, 10:30:34 pm
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hey i got the first and second parts correct but i cannot get the correct answer by my strategy for the third part about the vector equation. PLEASE HELP ME .
ANSWERS TO THE FIRST TWO PARTS :
ans1. DIM (range space)=3
ans2. basis of nullspace =(0, 1, -1, 1)
ans3. my strategy here was this:
Using the fact that if vector(b) is a member of the column space of and matrix(H) then there exists a vector(r), a member of the rowspace(H) which is a solution to the equation Ar=b, is unique and the magnitude(r) is the least possible, I first proved that (2, -10, -1, -15) is a member of column space of H then proved that (1, -3, 1, -2) is a member of the rowspace of H and hence the magnitude(1, -3, 1, -2) should be the least possible.
but the answer is in correct and i have no clue of any other approach to this question. please to me how to solve and also, kindly point out the whats wrong with my approach. I would be really thankfull if you explain me what a range space is as i feel that its a gray area.
thank you so very much.
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I will try and answer it tomorrow.
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Thanks.
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ans=sqrt(2^2+(-10)^2+(-1)^2+(-15)^2)
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No the answer is sqrt(3) and i cant see how
plz help me ...
thanks
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I will look again
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I know the method now. I made the nuull space (4a,a,-a,a) we can add mulitple of this to (1,-3,1,-2) and solve the equation H(1+4a,-3+a,1-a,-2+a)=(2,-10,-1,-15)
having done this use the given equation, d to obtain the square root of a quadratic in a. Find the minimum of this quadratic hence find the minimum distance.
Even for 7 marks quite a long question