Thread: First Year Mathematics A (Differentiation & Linear Algebra)

Originally Posted by Flop21
So knowing the definition of a continuous function, can you just say since the its lim = 0 from both the left and right hand side, it is continuous since as x->a f(x) = f(a).

Would writing just that be acceptable as an answer instead of using the eps-delta method?
Idk, by saying that the limits are 0 from the left and right, it seems (in my opinion) too close to assuming what they're asking you to prove.

They defined continuity for us somewhere along the lines of this

$\\Let f:[a,b]\rightarrow \mathbb R . Then f is continuous at c for some c\in [a,b] iff\\ \lim_{x\rightarrow{c}^{-}}{f(x)}=\lim_{x\rightarrow {c}^{+}}{f(x)} both exist and equal to f(c)$

(a) Show that if A is a 2 × 2 matrix, and its two rows are identical, then det(A) = 0.

(b) Prove by induction that if A is an n × n matrix which has two identical rows, then det(A) = 0.
[Hint: expand along one of the other rows].

How would one approach this question? Part a is quite easy, though I am not sure about part b.

Originally Posted by turntaker
(a) Show that if A is a 2 × 2 matrix, and its two rows are identical, then det(A) = 0.

(b) Prove by induction that if A is an n × n matrix which has two identical rows, then det(A) = 0.
[Hint: expand along one of the other rows].

How would one approach this question? Part a is quite easy, though I am not sure about part b.
|a al
|b bl = ab-ba

Originally Posted by iforgotmyname
|a al
|b bl = ab-ba
that's part a :P

Originally Posted by turntaker
(a) Show that if A is a 2 × 2 matrix, and its two rows are identical, then det(A) = 0.

(b) Prove by induction that if A is an n × n matrix which has two identical rows, then det(A) = 0.
[Hint: expand along one of the other rows].

How would one approach this question? Part a is quite easy, though I am not sure about part b.
$\noindent I'll sketch the idea (I assume you managed to do part (a)). Assume the result is true for n-1, and consider the n\times n matrix, which'll essentially be of this form: A=\begin{bmatrix} a_1 &a_2 &\cdots &a_n \\ a_1 & a_2 &\cdots &a_n \\ \vdots & \vdots & \ddots & \vdots\\ x_1 & x_2 &\cdots & x_n \end{bmatrix}.$

$\noindent (In general, the two identical rows could be in any of the rows, but as you know, we could always switch the rows to put them at the top, and the determinant would only possibly be changed in sign; so if we can prove that the matrix A above has zero determinant, it does it for the general case too.)$

$\noindent Consider expanding A along the last row. The determinant will look like x_j times the minor obtained by deleting the last row and column j, summed over j from 1 to n, i.e. \sum _{j=1} ^{n} (-1)^{n+j}x_j M_{n,j}, where M_{n,j} represents the minor we described. Now, each term of this sum will involve a determinant of a smaller matrix (size \left(n-1)\times \left(n-1\right)), but each of these determinants is just 0 from our inductive hypothesis, because if you look at those submatrices, they each contain the top two rows with one of the elements removed, so they still have the top two rows identical. It follows that the sum is 0, so \det \left(A\right)=0, as we wanted to show.$

(Parity of an integer refers to whether it is odd or even.)

Okay so with parametric vector forms of lines...

I was under the impression it was x = a + lambda(v). And I though V was simply AB (given two points A and B), so to get the vector you go B-A.

But in the solutions for my practice test it has A-B. Why have they got this, and if that is correct, why?

Originally Posted by Flop21
Okay so with parametric vector forms of lines...

I was under the impression it was x = a + lambda(v). And I though V was simply AB (given two points A and B), so to get the vector you go B-A.

But in the solutions for my practice test it has A-B. Why have they got this, and if that is correct, why?
Doesn't matter. Let mu = -lambda and donezo

Convention is to use B-A because A-B is technically a negative case

Originally Posted by Flop21
Okay so with parametric vector forms of lines...

I was under the impression it was x = a + lambda(v). And I though V was simply AB (given two points A and B), so to get the vector you go B-A.

But in the solutions for my practice test it has A-B. Why have they got this, and if that is correct, why?
$\noindent Yes, it can be the vector \overrightarrow{BA} or \overrightarrow{AB} or more generally any non-zero scalar multiple of \overrightarrow{AB}. This is because whichever non-zero scalar multiple of it we use, we'll end up covering precisely the same line as \lambda varies throughout the real numbers.$

Originally Posted by InteGrand
$\noindent Yes, it can be the vector \overrightarrow{BA} or \overrightarrow{AB} or more generally any non-zero scalar multiple of \overrightarrow{AB}. This is because whichever non-zero scalar multiple of it we use, we'll end up covering precisely the same line as \lambda varies throughout the real numbers.$
Oh right, because you can have multiple parametric vector forms for the same line? Because I was getting the signs of my vectors opposite to the answers (since they used A-B).

Originally Posted by Flop21
Oh right, because you can have multiple parametric vector forms for the same line? Because I was getting the signs of my vectors opposite to the answers (since they used A-B).
Yes, the parametric vector form for a given line is not unique. This is why we sometimes say "a" parametric form rather than "the" parametric form.

FML

I'm falling behind in maths. I need to catch up on lectures I've missed (due to catching up), but have trouble attending lectures since I don't understand what's going on because I've missed lectures / haven't had time to revise and understand.

So little time for anything tbh.

It's like a horrible cycle that I can't get out of, any advice anyone?

Originally Posted by Flop21
FML

I'm falling behind in maths. I need to catch up on lectures I've missed (due to catching up), but have trouble attending lectures since I don't understand what's going on because I've missed lectures / haven't had time to revise and understand.

So little time for anything tbh.

It's like a horrible cycle that I can't get out of, any advice anyone?
Surely you have lecture recordings on moodle

But the course pack does a reasonable job at filling people in if you ask me

Originally Posted by leehuan
Surely you have lecture recordings on moodle

But the course pack does a reasonable job at filling people in if you ask me
So I should skip this week's lectures as well and spend it catching up? I find myself prioritising studying for tests, so lately I've been just catching up on the upcoming test's material and not wanting to go ahead - but of course class is still speeding along with new crap while I study.

Course pack is really good. Man can't wait to start over next sem, I've learned so much, but kind have stuffed this sem.

Originally Posted by Flop21
So I should skip this week's lectures as well and spend it catching up? I find myself prioritising studying for tests, so lately I've been just catching up on the upcoming test's material and not wanting to go ahead - but of course class is still speeding along with new crap while I study.

Course pack is really good. Man can't wait to start over next sem, I've learned so much, but kind have stuffed this sem.
In advance for the calculus test you probably want to focus on studying for that. Same for algebra. Then when no calculus test is coming up you should catch up on the calculus that you missed.

But maths, well I view it as two seperate courses in algebra and calculus. You treat them seperate until final exam time.

As for missing, well that's probably up to you

Originally Posted by Flop21
FML

I'm falling behind in maths. I need to catch up on lectures I've missed (due to catching up), but have trouble attending lectures since I don't understand what's going on because I've missed lectures / haven't had time to revise and understand.

So little time for anything tbh.

It's like a horrible cycle that I can't get out of, any advice anyone?
If it makes you feel any better, I'm in the exact same boat for my subjects now

And to compound it, my supervisor sent me an email, "Hey SD, how's the work going? Let's meet next week to discuss your project"

catch is: i haven't done any work

she's going to kill me

Why is it when given some points after finding the cross product of vectors AB and AC, you find the result distance to find the area of the parallelogram?

How does that give you the area of a parallelogram?

Originally Posted by Flop21
Why is it when given some points after finding the cross product of vectors AB and AC, you find the result distance to find the area of the parallelogram?

How does that give you the area of a parallelogram?
$\noindent Recall that the length of the cross product of two vectors \vec{a} and \vec{b} in \mathbb{R}^3 is given by \left \| \vec{a} \times \vec{b} \right \| = \left \|\vec{a} \right \| \left \| \vec{b} \right \| \sin \theta, where \theta is the angle between the vectors. This is precisely the area of a parallelogram with side lengths \left \| \vec{a}\right \| and \left \|\vec{b} \right \| with angle \theta, as simple trigonometry shows us that \left \|\vec{a}\right \|\sin \theta is the perpendicular height of the parallelogram. See the diagram in this link:$

http://mathinsight.org/cross_product.

Okay can someone help me with finding the shortest distance d between a line and a point using vectors/projections??

These questions come up a lot and I'm having trouble doing them, so if anyone could give me a general list of steps of what to do to solve these, that would be great.

So say you get given a point - and it's said the line goes through that point, and is also parallel to a vector given, and you also get another point. They require you to find the projv/vector(the vector made by second point - first point).

Now what do you do from here to find the shortest distance between the line and the second point??

[if it would be easier for me to give a real example let me know]

I reckon draw a diagram and that'll help you out a lot - are you drawing pictures?

Yeah but it's always so hard in 3D to actually look at the diagram and have it useful

Originally Posted by Flop21
Yeah but it's always so hard in 3D to actually look at the diagram and have it useful
It suffices to draw a 2D diagram for that, even if the point and line are in some arbitrary n-dimensional space.

Originally Posted by InteGrand
It suffices to draw a 2D diagram for that, even if the point and line are in some arbitrary n-dimensional space.
do you mean drop the z number and just draw x and y?

Originally Posted by Flop21
do you mean drop the z number and just draw x and y?
We just need to draw a rough diagram to see what projection vector to take and what lengths to use etc. So we don't need to worry about coordinates for our sketch. Just draw a line with the known point on the line, and the given point above the line, and see which projection vectors and lengths need to be taken.

So basically pictures like in this link: https://en.wikipedia.org/wiki/Vector_projection .

If in 2014, the Australias population was 23 246 692, with a growth rate of approximately 1.14%, what was the population in 2015?

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