Maths 3A/3B

[spurcher]

HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      HERE TO HELP      Currently in yr12 and just about finishing;

Last year, in year 11 I studied this subject and finishes on 96. I am willing to help anyone with whatever question they may have and am willing to provide any extra material or links to external sites to get material from as well.......

Just saying Here to help anyone who might want/need it....

xoxo

[anotherworld2b]

Can i get help with c and d?

[jamonwindeyer]

Can i get help with c and d?

Hey! D isn't in your image!

For C though, it will be the inverse matrix of the one given for quadrilateral 1 to quadrilateral 2! That's the great thing about matrices representing transformations, to go backwards you just use the inverse matrix So it will be your typical 2x2 matrix inverse formula:

[anotherworld2b]

Thank you very much 
I also wanted to ask how do we prove matrices?

Hey! D isn't in your image!

For C though, it will be the inverse matrix of the one given for quadrilateral 1 to quadrilateral 2! That's the great thing about matrices representing transformations, to go backwards you just use the inverse matrix So it will be your typical 2x2 matrix inverse formula:

[jamonwindeyer]

Thank you very much 
I also wanted to ask how do we prove matrices?

For the question you provided, you would need to apply each transformation matrix (in order) to a general point, \(\binom{a}{b}\). So, take the general point, and apply each transformation by multiplying by the appropriate matrix. By the end, you should get back to the same point, \(\binom{a}{b}\).

[anotherworld2b]

Thank you for your help 
For matrices im confused in how to calculate the area. I followed the formula for the area of a triangle and i got 4 but the answer is 8?

Also i keep getting notifications that my images are too big and do not pass security checks?

For the question you provided, you would need to apply each transformation matrix (in order) to a general point, \(\binom{a}{b}\). So, take the general point, and apply each transformation by multiplying by the appropriate matrix. By the end, you should get back to the same point, \(\binom{a}{b}\).

[jamonwindeyer]

Thank you for your help 
For matrices im confused in how to calculate the area. I followed the formula for the area of a triangle and i got 4 but the answer is 8?

Also i keep getting notifications that my images are too big and do not pass security checks?

I'm not quite sure what you mean by the "area" of a matrix, could you elaborate?

I'm not sure about your images, what type of image are they? You could compress the files if they are too big maybe? 

[anotherworld2b]

Im having a lot of trouble posting images. A notification keeps telling me to consult the forum administrator for security checks?

I'm not quite sure what you mean by the "area" of a matrix, could you elaborate?

I'm not sure about your images, what type of image are they? You could compress the files if they are too big maybe? 

[jamonwindeyer]

Im having a lot of trouble posting images. A notification keeps telling me to consult the forum administrator for security checks?

No idea what that could be, I'll pass it on, but whatever is letting you post the images you do post, just keep doing that! It could be a weird file extension or something

I've never seen the idea of using matrices to find area, but a quick Google search yielded a few things? Maybe one of them will help you

[anotherworld2b]

Heres the question ive been working on. I finally succeeded at posting it 

[jamonwindeyer]

Heres the question ive been working on. I finally succeeded at posting it 

Awesome! I think you've misinterpreted a bit: ABCD is a square, not a triangle. Check the vertices and the question to confirm

So your first sketch needs fixing, and then the area formula you need would change in response to that. Let me know how that goes.

Using the determinant to find the area: I've never seen that before, so I'll have to leave you with it

I think the misinterpretation of using a triangle instead of a square is causing your issues

[anotherworld2b]

Could I get help with these two questions?

[jamonwindeyer]

Could I get help with these two questions?

Hey! Okay, so for Part A it's about finding the right way to consider the initial domains generally as a vector, then doing the transformation on the general point, thus proving the result for all points in the domain.

For example, we can express any point on the line \(y=5-3x\) as the following vector:



Apply the transformation to this vector, what you'll notice by doing the multiplication is that the x's cancel!



So the point in question is (10,5)! You'll do a similar thing for Part B, just consider a general point \(\binom{x}{y}\), and you'll find it maps to something that only has x's in it: This is a line!

Your second question, again the same principle. If we want to consider the transformation of the line, we can just consider the transformation of a general vector representing any point on the line!



Apply the matrix transformation to the vector, and you should get ANOTHER line with a new gradient, make the comparison as required if you have trouble snap a pic of your working and give me a look and I'd be happy to give more of a hand!

[anotherworld2b]

I'm still confused about how to do q21. I also tried to other questions 15 qne 16 but im not getting the right answer. I also wanted to ask for q17 what exactly is it asking?

Hey! Okay, so for Part A it's about finding the right way to consider the initial domains generally as a vector, then doing the transformation on the general point, thus proving the result for all points in the domain.

For example, we can express any point on the line \(y=5-3x\) as the following vector:



Apply the transformation to this vector, what you'll notice by doing the multiplication is that the x's cancel!



So the point in question is (10,5)! You'll do a similar thing for Part B, just consider a general point \(\binom{x}{y}\), and you'll find it maps to something that only has x's in it: This is a line!

Your second question, again the same principle. If we want to consider the transformation of the line, we can just consider the transformation of a general vector representing any point on the line!



Apply the matrix transformation to the vector, and you should get ANOTHER line with a new gradient, make the comparison as required if you have trouble snap a pic of your working and give me a look and I'd be happy to give more of a hand!

[jamonwindeyer]

I'm still confused about how to do q21. I also tried to other questions 15 qne 16 but im not getting the right answer. I also wanted to ask for q17 what exactly is it asking?

Where are you up to with Q21? Have you started by multiplying the vector I suggested above with the matrix given? That will apply the linear transformation, did you get that or is that operation troubling you?

Question 15 isn't really attackable with the method you used, you are better off using simultaneous!



Using the statement given, we conclude the following:



Do a similar thing for the multiplication of BA (the other way around), then you'll have four sets of simultaneous equations. Solve each to obtain an answer

For Q16 ,you have the right idea, but remember that matrix multiplication is not commutative! That is:



When you factored, you put the P out the front, it should have been out the back to preserve the initial order of the matrices, try again with:



Question 21 is asking you to consider general vectors and do a general proof of the statements given, kind of like a standard algebraic proof! so, consider A and B as non-singular (invertible) square matrices where \(AB=BA\), and prove generally that:



Let me know how you go!