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| 09-08-2012, 00:17 | #2 |
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Is it MA313 or MA314?
MA314 looks like a bit of a pain, but MA313 might be manageable. You could always post a question or two up here and we could have a lash at them. If you can ace the paper you failed, you should definitely pass the repeat. |
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| 09-08-2012, 10:08 | #3 |
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its MA313, the paper was a bit different this year from other years but i presume the repeat Will be very like the xmas paper! so yes all i need is the solutions really and i should be ok! are you in galway and available for a grind session by any chance? i Will post some of the questions up here now!
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| 09-08-2012, 10:37 | #4 |
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I don't give grinds, no, but you could do worse than to tell us which questions you can't do yourself. There are definitely a couple of posters on here who should be able to help with some of the questions, I suspect. Or you could try in the Mathematics forum: http://www.boards.ie/vbulletin/forumdisplay.php?f=380
Boards doesn't have a math font setting, so it might be easier to just say which questions you can't do and direct people to the downloadable pdf of the exam . |
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| 09-08-2012, 10:55 | #5 |
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q1. (a) For which value of k does the vector (1,2,k) belong to the linear span of {(1,1,1), (2,1,1)} ?
(b) For each of the following decide whether U is a linear sub space of the vector space V. Justify your answers with a short explanation in each case. (i) V = R3 and U = {(x,y,z) : x+y+z = 0} (ii) V is the space of all 3x3 matrices with real entries and U is the set of invertible 3x3 matrices. (iii) V is the space of all single variable polynomial functions of degree at most 2 and U = {f € V : f(1) = 0}. (c) show that the set {(1,1,1), (0,1,1), (2,1,0)} is linearly independent. |
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| 09-08-2012, 11:05 | #6 |
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| 09-08-2012, 11:29 | #7 | ||
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Quote:
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ie. So you need to decide what combination of (1,1,1) and (2,1,1) you could use to get (1,2,k). Well, [the second one] - [the first one] will give you the 1 you want for the first entry, but a 0 in the second entry so that's no good. How about [three times the first one] - [the second one]? That gives a 1 in the first entry, as required. And also a two for the second entry! Perfect. So the value it gives for the third entry, k, is 3-1=2. Alternatively: (1,2,k)=a(1,1,1)+(2,1,1)b. a+2b=1 a+b=2 k+1=b gives k=2. I'm sure you've looked at the paper and there are some questions which you can do, right? |
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| 09-08-2012, 17:08 | #9 | ||
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But you are, presumably, doing something, and can do some of the questions, which there's no need for other people to then spend their time trying to explain. A simple google of 3b, for example, yields this pdf which begins with the proof you're looking for for 3b.
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a) the zero vector of V is also in U b) that for any two vectors u1,u2 in U we have u1+u2 is in U. c) for any scalar k we have k*u1 is in U. If any of these conditions fail, it is not a subspace. So (i)(a) The zero vector of V is (0,0,0). Is this the zero vector in U? Yes, because x+y+z=0 as 0+0+0=0. (b) Choose two vectors in U, say u1=(x1,y1,z1) and u2=(x2,y2,z2). Is u1+u2 in U? u1+u2=(x1+y1+z1)+(x2+y2+z2)=(x1+x2,y1+y2,z1+z2). Does this satisfy the condition for U that x+y+z=0? Yes, because x1+x2+y1+y2+z1+z2=0. (We know this because u1 being in U tells us that x1+y1+z1=0. Similarly for u2. Then 0+0=0.) (c) If u1 is in U, is k*u1 in U? If u1=(x1,y1,z1) then k*u1=(kx1,ky1,kz1). Is (kx1,ky1,kz1) in U though? If it is, (kx1+ky1+kz1) would have to equal 0 to satisfy the condition on U. kx1+ky1+kz1=k(x1+y1+z1). And we know x1+y1+z1=0, because u1 is in U. Then k(x1+y1+z1)=k*(0)=0, so yes, k*u1 is in U. Since all three "checks" are satisfied, we know U is a subspace of V. Apologies for the crappy notation. That's the idea behind checking if something is a subspace. Quote:
1c A set of vectors x, y, z are linearly dependent if and only if there exist L1,L2,L3, (not all of which can be zero!) such that xL1+yL2+zL3 =0. So, supposing (1,1,1)L1+(0,1,1)L2+(2,1,0)L3=(0,0,0) this gives L1+2L3=0 L1+L2+L3=0 L1+L2=0 Solving this (I presume you can solve simultaneous equations) gives L3=0,L1=0, and L2=0. Since these must ALL be zero for L1,L2,L3 to exist, the vectors are NOT linearly dependent, meaning they are linearly independent. Last edited by Ficheall; 09-08-2012 at 17:34. |
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| 09-08-2012, 19:48 | #10 |
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Hmm...
For 2ai) you row-reduce and the rank is the minimum number of rows. aii) http://en.wikipedia.org/wiki/Column_space I'm not sure what they're looking for in b). Will have a proper look later. Your timing is terrible. You do know chess is on tonight, right? The answer for 2c is here, first definition: http://tutorial.math.lamar.edu/Class...geOfBasis.aspx |
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