The Student Room Group

Scroll to see replies

Hey all I've got some difficulty with one of the matrices question in a past paper question

SQA Past Paper 2007


It's question 5a only really, I'll take any inputs towards this question since I've tried looking over my notes, scholar and still don't know how to do it :/

Thank You

Sorry couldn't upload the question only because I'm on my iPad :tongue:
Original post by SameerM
Hey all I've got some difficulty with one of the matrices question in a past paper question

SQA Past Paper 2007


It's question 5a only really, I'll take any inputs towards this question since I've tried looking over my notes, scholar and still don't know how to do it :/

Thank You

Sorry couldn't upload the question only because I'm on my iPad :tongue:


When you multiply matrix A with i rows and j columns with matrix B with j rows and k columns it becomes an i * k matrix. Each element at position row-column i-k in the product matrix AB is the scalar product of the i-th row in A with the k-th column in B.

So in the question, the 2nd row of A is [0 1 -1]
The 3rd column of B is:
[x+3
2
3]

Their scalar product is 0 * (x+3) + 1 * 2 - 1 * 3 = -1. Therefore the element at 2nd row 3rd column of AB is -1. Repeat for elements at each position.
Thank you my friend, you have cleared out my difficulties in solving matrices :smile:
And you from Glasgow :redface: what year in uni are you in and which high school did you come from ?
Original post by SameerM
Thank you my friend, you have cleared out my difficulties in solving matrices :smile:
And you from Glasgow :redface: what year in uni are you in and which high school did you come from ?


Don't forget to ask him for his credit card details while you're at it.
Haha no I'm just curious since I'm from Glasgow and also because he's in Cambridge :smile:
Original post by SameerM
Haha no I'm just curious since I'm from Glasgow and also because he's in Cambridge :smile:


I'll tell you in PM.
Got my results back on my extended Unit 1 NAB - 66% or 21/32. I think I'm alright with that, ha. Obviously wish I'd done better but meh, could've been way worse :smile:
Passed my NAB! Full marks *phew* and 35/40 in extended test (20/20 in Algebra, 15/20 in Calculus.... made some really silly mistakes like forgetting to integrate the 2nd part of the equation :facepalm:)
Finally understand complex equations!
Reply 389
Original post by Mr Dangermouse
Finally understand complex equations!


:biggrin:

I like complex numbers. Moving on to integration on Thursday, so that's going to be fun (boring) since I've already done it all.
Not been much talk on the thread guys!

What topic is everyone on now? Liking it? Disliking it?

I'm currently doing further integration. Well almost done with complex numbers. Calculus is getting tricky but I really like all the algebra stuff.
(edited 12 years ago)
We're doing stuff in such a funky order. We started Unit 2 with further differentiation (we're done with that) and we're now in the middle of further integration AND complex numbers. Plus I've done proof and elementary number theory but no one else has, and I think a few people have done the stuff on sequences and series as preparation for the Oxford PAT. So... :P
Reply 392
Haven't seen this thread before. If y'all need help I should be as qualified(if not as clear!) as anybody to help if there's troubles with any particular sections. How I miss AH maths though!
Reply 393
Anyone know an actual use of complex numbers? We have just finished them and I can't seem to find a proper use for them. Ok so they let you find "non-real roots of polynomial equations". But doesn't that defeat the purpose of a root? Shouldn't it be where it crosses the x-axis i.e. when y=0. If it doesn't cross it then it doesn't have any roots not these "complex roots".
Reply 394
Original post by soup
Anyone know an actual use of complex numbers? We have just finished them and I can't seem to find a proper use for them. Ok so they let you find "non-real roots of polynomial equations". But doesn't that defeat the purpose of a root? Shouldn't it be where it crosses the x-axis i.e. when y=0. If it doesn't cross it then it doesn't have any roots not these "complex roots".


Essentially the complex numbers are just an extension of our number system. Have you heard of diophantine equations? They're equations where solutions can only be integers, eg 3x=2 has no solutions. Or in the past, before negative numbers were considered ok the equation x+1=0 would be considered insoluble.
Equally, when we only have rationals(fractions), we would say the equation x^2=2 is insoluble. Moving from the positive integers, to all integers, to rationals to reals is just adding to the number of polynomials we can totally solve.
In adding the complex numbers, we get to a complete field, that is every polynomial of degree n will now have n roots. That's basically the point of them, and leads onto the Fundamental Theorem of Algebra.
Original post by soup
Anyone know an actual use of complex numbers? We have just finished them and I can't seem to find a proper use for them. Ok so they let you find "non-real roots of polynomial equations". But doesn't that defeat the purpose of a root? Shouldn't it be where it crosses the x-axis i.e. when y=0. If it doesn't cross it then it doesn't have any roots not these "complex roots".


Well, we've only had one lesson on them, but our teacher mentioned in passing that the maths of complex numbers underlies a lot of physics about electricity. Don't ask me how :L

The whole point of having a new number system is that the solutions are a different thing, but still useful. When negative integers were first used, people might have questioned the value of a solution to x+4=3x+4=3, since from a counting point of view you can't have 1-1 apples. But negative integers have their uses both in maths and the sciences, and the same is true of complex numbers - even if a complex solution isn't quite the same thing as a real solution.
Reply 396
Original post by Slumpy
Essentially the complex numbers are just an extension of our number system. Have you heard of diophantine equations? They're equations where solutions can only be integers, eg 3x=2 has no solutions. Or in the past, before negative numbers were considered ok the equation x+1=0 would be considered insoluble.
Equally, when we only have rationals(fractions), we would say the equation x^2=2 is insoluble. Moving from the positive integers, to all integers, to rationals to reals is just adding to the number of polynomials we can totally solve.
In adding the complex numbers, we get to a complete field, that is every polynomial of degree n will now have n roots. That's basically the point of them, and leads onto the Fundamental Theorem of Algebra.


Original post by derangedyoshi
Well, we've only had one lesson on them, but our teacher mentioned in passing that the maths of complex numbers underlies a lot of physics about electricity. Don't ask me how :L

The whole point of having a new number system is that the solutions are a different thing, but still useful. When negative integers were first used, people might have questioned the value of a solution to x+4=3, since from a counting point of view you can't have -1 apples. But negative integers have their uses both in maths and the sciences, and the same is true of complex numbers - even if a complex solution isn't quite the same thing as a real solution.


I think the problem is that we have been introduced to complex numbers and taught no real-world applications for them.
Reply 397
Original post by soup
I think the problem is that we have been introduced to complex numbers and taught no real-world applications for them.


There aren't any that immediately come to mind for me. As has been said, complex numbers underpin most of electronics, but I don't really have enough background to expand upon that.
Edit-also, are you being taught applications for most of the course?
Original post by soup
Anyone know an actual use of complex numbers? We have just finished them and I can't seem to find a proper use for them. Ok so they let you find "non-real roots of polynomial equations". But doesn't that defeat the purpose of a root? Shouldn't it be where it crosses the x-axis i.e. when y=0. If it doesn't cross it then it doesn't have any roots not these "complex roots".


That is because for the interpretation of "polynomial equations" as a line on the x-y plane to be valid, you are implicitly restricting the function to the domain of real numbers, like you are taught to do through school before AH. Once you admit complex solutions, you are actually expanding the domain and codomain to complex numbers and so the naiive representation as a line graph over the x-y plane is not valid any more. You are now dealing with functions that can be expressed as f(x + yi) = a + bi. To represent this adequately, you need a complex surface over the plane like this:

The x-y position in the graph corresponds to x and y in the input to f. The brightness of a point is the magnitude of the result a + bi and the colour is determined by the argument of a+ bi. Hence complex roots, where a = b = 0, appears as points of black on this surface (look closely at the tips of the each colour band).

So it's not that these complex roots don't exist or don't make sense, it's just that the traditional representation of a function is not adequate enough to interpret them.
(edited 12 years ago)
Reply 399
Original post by soup
Anyone know an actual use of complex numbers? We have just finished them and I can't seem to find a proper use for them. Ok so they let you find "non-real roots of polynomial equations". But doesn't that defeat the purpose of a root? Shouldn't it be where it crosses the x-axis i.e. when y=0. If it doesn't cross it then it doesn't have any roots not these "complex roots".


Used in Quantum Mechanics a lot. The wavefunctions have imaginary numbers in them a lot of the time, but no one really knows what an imaginary number means in a physical sense; this is usually shortcutted by squaring the wavefunction. I believe there's a lot to do with complex conjugates and stuff too. You'd be better asking a physics, maths or chemistry student for further explanation about the QM stuff.

Latest

Trending

Trending