how damn hard is kilbaha 2010 Exam 1?Tech Free

[TyErd]

in question 1, how do you find out 'Q'. I know how to figure out 'P'.

[8039]

in question 1, how do you find out 'Q'. I know how to figure out 'P'.

Did you arrange a matrix? then get the determinent?

[TyErd]

yea

[Martoman]

Inverse question is a bitch.

AAANNNDDD the triangle increasing degree one can be annoying if you don't understand trig differentiation and the reason for it being in radians.

[8039]

yea

well I got q first by just doing a typical simultaneous equation

[TyErd]

but how with simultaneous equations do you get an answer of q e R

[Martoman]

but how with simultaneous equations do you get an answer of q e R

For this I used a bit of sense. I didn't do any matricies rubbish.

For the thing to have a unique solution they must not have the same gradient. So find the ratio of the components to see what value of p WOULD give the same gradient.



if they have the same gradient. IF they have the same gradient then we know that a unique solution doesn't exist. So then p can take any value but 15/4 and for q it doesn't matter. It can be anything.

For the second part, we imagine two lines parallel. If they are parallel they will never intersect, but as with part iii they can coincide for one value of q only, which will lead to infinite solutions.

So to answer ii, i answered iii first. If they lie on the same line they will share points. So I found that on the second equation, x = 5 when y = 0 so there is a point shared at (5,0)

Now we know that to have infinite solutions p must be fixed at 15/4 so -3(5) = q so q = -15 and p = 15/4 for infinite solutions.

In contrast and now to solve part ii, for there to be no solution they are parallel but cannot lie on the same line. Thus p must be 15/4 and q can take any value but -15.

All you guys have to do to solve these kinds of questions is use some visualisation and the methods become quite trivial.  :smitten:

[wildareal]

Anyone have a copy, would be interesting to see it

[Martoman]

Anyone have a copy, would be interesting to see it

*points to cyberspace* http://vcenotes.com/forum/index.php/topic,29541.0.html

[8039]

but how with simultaneous equations do you get an answer of q e R

For this I used a bit of sense. I didn't do any matricies rubbish.

For the thing to have a unique solution they must not have the same gradient. So find the ratio of the components to see what value of p WOULD give the same gradient.



if they have the same gradient. IF they have the same gradient then we know that a unique solution doesn't exist. So then p can take any value but 15/4 and for q it doesn't matter. It can be anything.

For the second part, we imagine two lines parallel. If they are parallel they will never intersect, but as with part iii they can coincide for one value of q only, which will lead to infinite solutions.

So to answer ii, i answered iii first. If they lie on the same line they will share points. So I found that on the second equation, x = 5 when y = 0 so there is a point shared at (5,0)

Now we know that to have infinite solutions p must be fixed at 15/4 so -3(5) = q so q = -15 and p = 15/4 for infinite solutions.

In contrast and now to solve part ii, for there to be no solution they are parallel but cannot lie on the same line. Thus p must be 15/4 and q can take any value but -15.

All you guys have to do to solve these kinds of questions is use some visualisation and the methods become quite trivial.  :smitten:

Good explanation... but I wouldn't say matrices are rubbish. Especially when there is more than one value to the variable to have no solutions etc. it makes it much easier imo

[Martoman]

Oh, i'm sure its not, its just the reliance and robotic nature that it gets you thinking in when you can apply what I did in about 3-4 lines. Matricies and their applications are amazing without a doubt however.

[TyErd]

your explanations are brilliant Martoman, would you be kind enough to explain how to solve with matrices as well

[Martoman]

matricies? I would find p with the determinate = 0 so (-3)(-5) -4p = 0 if there is no solution so p = 15/4

for q I'd then apply my logic.

[the.watchman]

matricies? I would find p with the determinate = 0 so (-3)(-5) -4p = 0 if there is no solution so p = 15/4

for q I'd then apply my logic.

Ah but can be infinite or no solutions

[Martoman]

matricies? I would find p with the determinate = 0 so (-3)(-5) -4p = 0 if there is no solution so p = 15/4

for q I'd then apply my logic.

Ah but can be infinite or no solutions

Refer to my previous post. I meant it with reference to that logic.