A-Level Further Maths: C4-04 Invariance: Example of Finding Invariant Lines

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[@charliescoulding6324]

Been learning from home because of the pandemic and trying to catch up, couldn't understand it from my school powerpoint but this was really helpful

[@dingzhong9339]

Thank you so much sir, this is extremely helpful.

[@TLMaths]

Glad it was helpful!

[@obedientcape4912]

This is not in the core pure mathematics text book. But it has come up in previous papers so thank you.

[@boringblobking3783]

very tasty video

[@danuzwanus9334]

Extra tasty

[@gddanielk8491]

Scrumptious

[@puddleduck1405]

scrumdillydillydilcious

[@kusalyarodrigo1727]

Thank you so much for these videos.

[@TheObloINATOR777]

thanks so much! this really helped me before my finals 😁

[@casperlui5728]

I am really appreciate to this video

[@sameccleston8673]

If both y= 5/2x + c and y= -x are both invariant lines and the translated coordinate will remain on those lines, will the intersections between these invariant lines be invariant points, as that seems to me the only way the translated point can satisfy both invariant lines? Or am I missing something?

[@TLMaths]

Good question I hadn't thought about before. I would assume the answer to this is yes, as otherwise there would be points on an invariant line that do not remain on that line. So wherever two invariant lines intersect, that point must be an invariant point.

[@mukhtarquraishi4233]

Great Video!!!!!!! Thanks so much, it's very helpful

[@Theproofistrivial]

How does this work for matrices which don't provide an X and a Y term in both elements, such as a multiple of the identity matrix (kI)? You are left with kX equal to X' (and kY=Y') which means that when X' and Y' are substituted into y=mx + c, you get Y'=m(X') + c and hence kY=m(kx) + c and therefore kmx + kc = kmx +c which suggests m can be anything when c=0.

[@godfred8618]

amazing teacher

[@blahwoofyackety2563]

Thank you very very much! Really helpful and clear

[@GrettaLem]

how does this work when there is an identity matrix or an identity 'like' matrix e.g. swap out the 1 in the identity matrix for -3 as surely it will just get cancelled each time suggesting it has no invarient lines despite this not being the case?

[@BBQsquirrel]

Could this also be done using eigenvectors and eigenvalues?

[@TLMaths]

Yes, and that would be my go-to method in general.

[@brandon-D]

Great explanation, thank you

[@katiec5352]

what a legend no way this guy is amazing

[@danielmillward5842]

Im sorry TL maths, I love you dearly, but this is the worlds most complicated way of finding invariant lines.

[@TLMaths]

Happy to listen to an easier method that works in all cases.

[@chesster5981]

It’s a way that follows logically for me, just have you assume there exists a line y=mx+c

[@bigboyskrapz]

Why are we assume the coefficient of both x and c are 0 to solve for m. How do you know that it’s not 3x-4c=0 for example. Because that could still equal 0

[@TLMaths]

We know that 0 = (3m^2 - 3m - 5)x + (2m - 5)c, so by comparing coefficients, 3m^2 - 3m - 5 = 0 and (2m - 5)c = 0

[@musa6138]

at 3:17 isnt it supposed to be m[(3 + 2m) x + 2mc] because all of it is x'

[@TLMaths]

You've gained an extra m, it is: m[(3 + 2m) x + 2c]

[@musa6138]

TLMaths from having 0 = .... +( 2m^2 -5) c wouldn’t that give you m = (5/2)^(1/2)?

[@TLMaths]

It's (2m - 5)c, not (2m^2 - 5)c

[@williamhallleiva8735]

Thank you for this video, a really well explained video :)

[@moodymonstrosity1423]

Why we substituted mx+c into y' ?

[@TLMaths]

We assumed that the invariant lines are of the form y = mx + c, so we can replace y with mx + c. Once we have y' and x', we know that y' = mx' + c as the line that is outputted by the transformation must be the same line as what we started with (so same gradient m and y-intercept c).

[@moodymonstrosity1423]

@@TLMaths ok thankyou

[@JamesTaylor6]

Suddenly it's all so clear

[@jasoncho2753]

Does this method take into account enlargements? After all, lines of invariance can be created via enlargements as well

[@shaheerziya2631]

Thank you kind sir.

[@jamminermit]

I was watching a video about eigenvectors, and they seemed similar to invariant lines. Are they the same idea, and if so could you use eigenvectors to find the equation of these lines?

[@TLMaths]

Eigenvectors point in the direction of invariant lines. Eigenvalues are effectively the scale factor that tells you how far points have moved along those lines. So if the eigenvalue for a particular eigenvector is 1, then it is a line of invariant points.

[@quirkyzigzag]

wouldnt it be easier as this is under the linear transformations topic to just have y= mx and neglect the c as it should map to the origin if treated as a transformation. I can see this being useful if it it is not but for this topic ??

[@TLMaths]

Matrices transformations can leave lines that aren’t going through the origin invariant, such as a reflection matrix

[@awall007]

Perfect thanks

[@lucasf.v.n.4197]

another way to grasp eigenvalues and eigenvectors, I like this angle;