08-08-2012, 23:20 #1 Clare13 Registered User   Join Date: Aug 2012 Posts: 6 URGENT Maths Help! 3rd year maths student in need of help with linear algebra! Exam is on Friday afternoon so would need to meet tomorrow or Friday morning the latest! If anyone can help please let me know asap !
 09-08-2012, 00:17 #2 Ficheall Registered User   Join Date: Aug 2010 Posts: 3,334 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 Clare13 Registered User   Join Date: Aug 2012 Posts: 6 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!
 09-08-2012, 10:37 #4 Ficheall Registered User   Join Date: Aug 2010 Posts: 3,334 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 Clare13 Registered User   Join Date: Aug 2012 Posts: 6 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.
 09-08-2012, 11:05 #6 Clare13 Registered User   Join Date: Aug 2012 Posts: 6 https://www.mis.nuigalway.ie/regexam...ish_flag=False would really appreciate any solutions!
09-08-2012, 11:29   #7
Ficheall
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Join Date: Aug 2010
Posts: 3,334
Quote:
 Originally Posted by Clare13 https://www.mis.nuigalway.ie/regexam...ish_flag=False would really appreciate any solutions!
Ah now, come on. Surely you can do some of them. There's no point in people just doing out the questions if a) you've already done them, or b) you're not prepared to put some work in yourself. ANYONE on boards could memorise the winter paper and pass the exam on Friday. Where'd be the fun in doing the course if you didn't have to put in at least some mathematical effort?

Quote:
 Originally Posted by Clare13 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)} ?
Here you want to write (1,2,k) as some combination of the other two vectors and solve for k.
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, 15:44 #8 Clare13 Registered User   Join Date: Aug 2012 Posts: 6 . To be honest at this very moment in time I can do very little. Thanks for your help
09-08-2012, 17:08   #9
Ficheall
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Join Date: Aug 2010
Posts: 3,334
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.

Quote:
 Originally Posted by Clare13 (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}.
To verify U is a subspace of V, you need to verify that:
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:
 Originally Posted by Clare13 (c) show that the set {(1,1,1), (0,1,1), (2,1,0)} is linearly independent.

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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