WACE Stuff > Mathematics Stage 3
Maths 3A/3B
anotherworld2b:
I was able to get q11 right but i could get q13 right. For q14 i followed an example from our textbook but i atill didnt get the answer.
Thank you for your help I rewlly appreciate it.
--- Quote from: jamonwindeyer on September 07, 2016, 10:39:59 pm ---Oh! The matrix I gave you in my original explanation had a sign error for Q15, dictation error, sorry! Go back and look at your first line of working, do the matrix multiplication again, fix the sign errors and repeat, your method is 100% correct ;D
Question 13, best way to do it would (I think) just be considering a general matrix \(\begin{bmatrix}a & b\\c &d\end{bmatrix}\) and setting up simultaneous equations just like above!
Question 14, your attempt isn't quite correct. To make it clearer, try drawing a little diagram!
Triangle 1 ------ (Matrix A) -------> Triangle 2
Triangle 1 ------ (Matrix B) -------> Triangle 3
The idea here is to go from Triangle 2 to Triangle 3. We can do this by going back to Triangle 1, then to triangle 3, the matrix therefore being:
where A is the first matrix given, B is the second matrix given :)
Have you seen this method before? If not happy to explain! :)
Question 12 should be fairly straightforward, calculate the inverse:
Then multiply it by itself, form 4 equations and then solve for the missing variables! Show me your working if that doesn't quite work! :P
Then Q13 I addressed above ;D
Hope this helps!! ;D
--- End quote ---
jamonwindeyer:
--- Quote from: anotherworld2b on September 08, 2016, 01:07:23 am ---I was able to get q11 right but i could get q13 right. For q14 i followed an example from our textbook but i atill didnt get the answer.
Thank you for your help I rewlly appreciate it.
--- End quote ---
Q13A should be:
That's if you wanted to do it with two separate equations instead of one big one. Note that the second vector must be a column vector, otherwise the transformation doesn't make sense. Of course what we notice is that the answer is actually just formed with the columns of the vectors in the question, that's a consequence of transforming the standard basis vectors in \(\mathbb{R}^2\), but it's cool if you don't spot that ;D
Have another go ;D
As I said for 14, incorrect working, your worked example in the textbook is slightly different to your question. It's not the triangle \(P'Q'R'\) being transformed to \(P''Q''R''\), it is \(PQR\).
You need the method I showed you above (or something similar) :)
anotherworld2b:
oh :D thank you
I'm not sure how to do q14?
--- Quote from: jamonwindeyer on September 08, 2016, 09:36:39 am ---Q13A should be:
That's if you wanted to do it with two separate equations instead of one big one. Note that the second vector must be a column vector, otherwise the transformation doesn't make sense. Of course what we notice is that the answer is actually just formed with the columns of the vectors in the question, that's a consequence of transforming the standard basis vectors in \(\mathbb{R}^2\), but it's cool if you don't spot that ;D
Have another go ;D
As I said for 14, incorrect working, your worked example in the textbook is slightly different to your question. It's not the triangle \(P'Q'R'\) being transformed to \(P''Q''R''\), it is \(PQR\).
You need the method I showed you above (or something similar) :)
--- End quote ---
jamonwindeyer:
--- Quote from: jamonwindeyer on September 07, 2016, 10:39:59 pm ---Question 14, your attempt isn't quite correct. To make it clearer, try drawing a little diagram!
Triangle 1 ------ (Matrix A) -------> Triangle 2
Triangle 1 ------ (Matrix B) -------> Triangle 3
The idea here is to go from Triangle 2 to Triangle 3. We can do this by going back to Triangle 1, then to triangle 3, the matrix therefore being:
where A is the first matrix given, B is the second matrix given :)
--- End quote ---
This was the explanation I provided earlier. Again, the idea is that to go from Triangle 2 (P'Q'R') to Triangle 3 (P''Q''R''), we need to go back to Triangle 1 by multiplying by the inverse of the matrix given, THEN multiply (on the left) to get to Triangle 3. The product of those matrices must, by definition, be the same as the matrix we need, which is where that expression above comes from ;D
anotherworld2b:
Then would you do the same thing for Q13b?
--- Quote from: jamonwindeyer on September 08, 2016, 10:12:50 am ---This was the explanation I provided earlier. Again, the idea is that to go from Triangle 2 (P'Q'R') to Triangle 3 (P''Q''R''), we need to go back to Triangle 1 by multiplying by the inverse of the matrix given, THEN multiply (on the left) to get to Triangle 3. The product of those matrices must, by definition, be the same as the matrix we need, which is where that expression above comes from ;D
--- End quote ---
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