Problem Solving Strategies

Abcd

6:10 AM

@JohnRennie Are you there

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie hi, morning. How can a non signular matrix satisfy $AB - BA = A$

Because if we take determinant both sides we get $|A|= 0$

@JohnRennie Basically the question is if a non singular matrix satifises AB -BA = A then prove that det(B+I)= det(B-I)

But I doubt the validity of the question because of the aforementioned reason;

John Rennie

To be honest this is a bit beyond my grasp of matrix algebra. I can have a go at finding the answer, probably with Google's help, if you want but I can't give you the answer straight off.

Abcd

@JohnRennie Okay leave it.

John Rennie

6:29 AM

@Abcd $A^{-1} AB - A^{-1}BA = A^{-1}A$

So $B - I = A^{-1}BA$

Likewise $ABA^{-1} - BAA^{-1} = AA^{-1}$

So $B + I = ABA^{-1}$

So does $\det(A^{-1}BA) = \det(ABA^{-1})$ ?

The non-singular requirement is the requirement that A have an inverse

Aha, that's true isn't it, because the determinant of a matrix product is the product of the determinants.

$\det(A^{-1}BA) = \det(ABA^{-1}) = \det(B) $

Abcd

@JohnRennie yes why not

John Rennie

@Abcd so there you go, that's the proof $\det(B+I) =\det(B-I)$

Abcd

@JohnRennie my mistake was: $\det(A+B) \ne \det(A)+ \det(B)$

John Rennie

@Abcd that's true in general

Abcd

@JohnRennie no man its not

John Rennie

6:45 AM

@Abcd oops, I meant your statement is true in general

That is, I'm agreeing with you

Abcd

Oh okay.

Abcd

7:34 AM

@JohnRennie Are you there

John Rennie

@Abcd hi

Abcd

@JohnRennie I don't get what he's trying to say ^^^

John Rennie

It seems to be well described here:

Elementary matrix

In mathematics, an elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation. The elementary matrices generate the general linear group of invertible matrices. Left multiplication (pre-multiplication) by an elementary matrix represents elementary row operations, while right multiplication (post-multiplication) represents elementary column operations. Elementary row operations are used in Gaussian elimination to reduce a matrix to row echelon form. They are also used in Gauss-Jordan elimination to further reduce the matrix to reduced row echelon...

Or at least that covers theorem 1

Abcd

@JohnRennie where

John Rennie

The Elementary row operations section explains how you use a prefactor to achieve the transformation

Oh, wait, no it doesn't.

Oh yes it does

Abcd

7:42 AM

@JohnRennie Author's language is confusing$^\infty$ . Please tell me in simple words what he means.

John Rennie

If E is an elementary matrix, as described below, to apply the elementary row operation to a matrix A, one multiplies A by the elementary matrix on the left, E⋅A. The elementary matrix for any row operation is obtained by executing the operation on the identity matrix.

@Abcd there are only three elementary row operations.

Take the scaling one as an example because it's simple.

Abcd

Okay

??

John Rennie

Suppose you want to multiply the first row of a matrix by a factor $a$. Take I and multiply its first row by $a$.

Then premultiply your matrix by the modified identity.

Abcd

@JohnRennie But then I's first row will be like a 0 0 (for order 3)

John Rennie

Let me attempt some mathjax ...

$$ \left( \begin{matrix} X & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) = \left( \begin{matrix} Xa & Xb & Xc \\ d & e & f \\ g & h & i \end{matrix}\right) $$

So to multiply the first row by $X$ perform this operation on I to get:

$$ \left( \begin{matrix} X & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) $$

Then premultiply your matrix by your modified I matrix

And I think theorem 2 follows from theorem 1 using the associativity of matrix multiplication.

Abcd

8:05 AM

@JohnRennie ?? What do you mean here?

John Rennie

Suppose we have some matrix A and we want to multiply its first row by the constant X.

Abcd

So perform the transformation R1 -> xR1 in I then premultiply

Is that what you mean?

John Rennie

@Abcd yes

Abcd

@JohnRennie whats the use when you can just directly do it on original matrix !?

John Rennie

It's common in maths that you establish theorems because they can be useful later.

For example once we have theorem 1 then theorem 2 follows from it.

Abcd

8:08 AM

@JohnRennie Didn't get Theorem 2 please explain that as well.

John Rennie

So the point of theorem 1 is not to give you a faster/better/whatever way to do an elementary transformation, but to provide a tool for working with more complicated calculations.

@Abcd OK, suppose we want to do an elementary row transformation on the product AB - I don't know what the notation for this is but let's use $X(AB)$ to mean the elementary transformation $X$ applied to $AB$

Abcd

@JohnRennie elementary transformation is denoted by $\ce{R_1 -> XR_1}$ right?? if you want to multiply row 1 by x

John Rennie

@Abcd yes, that's an example of an elementary transformation.

Abcd

@JohnRennie ok then?

John Rennie

Theorem 1 tells us that $X(AB) = X(I)(AB)$. Yes?

Abcd

8:12 AM

yes

Oh

so:

$(XA)B$

John Rennie

BOOM!! :-)

Now you see why theorem 1 is useful, even if it seemed a bit pointless at first.

Abcd

Okay thanks!

@JohnRennie But why does he say its also valid for post factor case?

Oh postfactor for column thing

Okay got it!

John Rennie

So far we've talked about row transformations.

The same approach applies to column transformations, but you have to postmultiply by your transformed identity matrix, not premultiply.

Abcd

yeah

John Rennie

If you use my example above it should be obvious that this works.

Abcd

8:18 AM

@JohnRennie Please show row addition elementary transformation theorem 1 proof just like you showed for row multiplication

John Rennie

OK, suppose you want to set row 1 to be the sum of rows 1 and 2. Is that a good example to take?

Abcd

Just a minute. Let me recheck my work I think Ive made a mistake.

okay done.

Proved myself. It works. Thanks.

John Rennie

Cool :-)

harambe

9:11 AM

@JohnRennie good evening

John Rennie

@harambe hi :-)

Nehal Samee

10:25 AM

@JohnRennie ...there?

John Rennie

@NehalSamee hi

Nehal Samee

@JohnRennie Hi ....

I'd a problem...

John Rennie

@NehalSamee yes?

Nehal Samee

A lens is placed such that it produces a sharp image. If right half of lens is made opaque, then what would happen to image?

@JohnRennie Concerning its brightness, magnification and any possible property ...

John Rennie

Any lens is a combination of a focussing device (the lens) and an aperture through which the light passes. For a regular circular lens the aperture is just the area covered by the lens.

The image created by this is a combination of a perfect image combined with the Fourier transform of the aperture.

Nehal Samee

10:30 AM

@JohnRennie So?

John Rennie

For a regular circular lens the image is a perfect image convolved with an Airy disk, which is the Fourier transform of a circular aperture.

Under most circumstances the size of the Airy disk is far smaller than the size of the image so at worst all this does is cause a very slight blurring of the image.

Nehal Samee

@JohnRennie Will it resume magnification to half of earlier?

John Rennie

In your example you have a semicircular aperture, so the image will be convolved with whatever the Fourier transform of a semicircle is. Offhand I don't know what it is.

Nehal Samee

@JohnRennie OK 😶

John Rennie

@NehalSamee the image remains the same, but it is convolved with a different Fourier transform. However under most circumstances this will have little effect on the image.

Nehal Samee

10:35 AM

@JohnRennie OK...

John Rennie

However because you've covered half the lens only half as much light gets through. So the main effect is that the image is only half as bright.

Nehal Samee

@JohnRennie Got it...

John Rennie

@NehalSamee in general masking out bits of a lens has surprisingly little effect on the image except to make it less bright.

Nehal Samee

@JohnRennie OK.

harambe

1:28 PM

@sammygerbil good afternoon! I wanted to tell you I got the solution thanks to your method

sammy gerbil

1:46 PM

@harambe well done. It is a difficult problem, but solvable if you are careful.

Nehal Samee

2:38 PM

@sammygerbil could you see a problem ?

Two bodies of same mass m rotate around each other (like binary stars) at a distance r from each other. Show that, each has a velocity of $v=1/2*\sqrt{Gm}{r}$

sammy gerbil

@NehalSamee Have you tried to solve this? Are you stuck?

The question says the bodies rotate around each other, but in fact they rotate around their COM, which is at their midpoint.

So the radius of their orbit $r$ is not the same as the distance between them $R=2r$.

The centripetal force on each body is $mv^2/r$ whereas the gravitational force of attraction is $Gmm/R^2$.

Nehal Samee

5:58 PM

@sammygerbil I did this the same way as you said... But I got 1/√2 instead of 1/2

sammy gerbil

@NehalSamee Can you explain how you did it?

Nehal Samee

@sammygerbil Equated 2mv^2/r and Gmm/r^2 , where r is distance between them...

sammy gerbil

@NehalSamee Where did the 2 come from? Is that because there are 2 bodies?

Nehal Samee

@sammygerbil No... That's only for a particular body... r/2 is what I used...

sammy gerbil

You should have $mv^2/r=Gmm/R^2$.

Nehal Samee

6:05 PM

@sammygerbil In my equation, r=R/2

Is the expressiom given in question to prove, correct?

sammy gerbil

Ah yes sorry that is my fault. No the expression is not correct, if $r$ is the distance between the bodies.

Nehal Samee

@sammygerbil OK got it... Thanks for clarifying my doubt...

sammy gerbil

Probably the question-setter also got confused between radius of orbit $r$ and separation between bodies $R=2r$.

@NehalSamee So you had the correct calculation all the time?

Mr. Xcoder

8:15 PM

Hi! I have very little calculus experience and the following came across in one of my exercises: area of circular sector (without using circle formula)... More like mathematical-physics, but can anyone check if my reasoning is good: $$A=\int \mathrm{d}A\implies A=2\int_0^{\alpha/2}R^2\mathrm{d}\theta/2=\\=R^2\int_0^{\alpha/2}\mathrm{d}\theta‌=R^2\alpha/2$$ Where $\alpha$ is in radians. It does check off for the full circle and its fractions so I guess it's fine, but is it rigorous and correct?

Where $\alpha$ is the central angle of the circular sector, $R$ is its radius and $\mathrm{d}\theta$ is an infinitesimal change in the angle. Method used: cutting sector into infinitesimally small triangular pieces

I am particularly doubtful about my very first equality, it seems... Ugh IDK, strange ($A=\int \mathrm{d}A$)

sammy gerbil

@Mr.Xcoder Nothing wrong with that. $dA$ and the limits are yet to be defined.

It gives the correct answer, and you can verify the formula on the internet, so I think you are being too cautious.

Mr. Xcoder

I guess so, you are likely right. But I thought it's quite strange because it basically proves the circle area formula... in 2 lines? Anyway, I guess it's just a consequence of the power of calculus.

Thanks!

I was also more cautious than usual because although it gives the same answer as the Wikipedia entry, that one uses a more sophisticated integral

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Problem Solving Strategies