AQA FP4 25th May 2016 Discussion Thread

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What's it actually mean when a set of equations are said to be inconsistent or consistent ?
I don't understand the meaning of that :/

Reply 101

Argylesocksrox

Original post by Roxanne18

What's it actually mean when a set of equations are said to be inconsistent or consistent ?
I don't understand the meaning of that :/

Say you have equation x+y=3 and x+y=2, it is obvious that these equations cannot be solved (IE have no point of intersection), if you solve for x/y you get the contradiction 2=3, which is obviously wrong.

works for x,y,z etc.

Reply 102

Argylesocksrox

I'm probably going to have problems sleeping tonight, does anyone have any recommendations? (Don't say be calm, I struggle with being calm)

Reply 103

C0balt

Original post by Argylesocksrox

I'm probably going to have problems sleeping tonight, does anyone have any recommendations? (Don't say be calm, I struggle with being calm)

Been suffering from insomnia past couple of months and I was recommended to do progressive muscle relaxation exercise and mindfulness breathing which you can look up on YouTube. Whether it's effective or not, I'm not sure, it might be helping me but not enough lol

Reply 104

Argylesocksrox

Original post by C0balt

I'll try the breathing, and maybe a jog would be beneficial.

Reply 105

B_9710

Here's an extension question if anyone wants to be sure about solving systems of equations.

\displaystyle a+2b+c+d=7

\displaystyle a+2b+2c-d=12

\displaystyle 2a+4b+6d=4

.

Find the general solution to the system of equations.

Reply 106

Qcomber

i can get that c=2t+5 and d=t.
but cant find a and b

Reply 107

WillTheWiener

Can anyone tell me how you would work out invariant lines or lines of invariant points from eigenvalues? I believe that when λ = 1 you get a line of invariant points but what about when λ ≠ 1?

Reply 108

Qcomber

Original post by WillTheWiener

Can anyone tell me how you would work out invariant lines or lines of invariant points from eigenvalues? I believe that when λ = 1 you get a line of invariant points but what about when λ ≠ 1?

The eigenvector represents the direction vector for an invariant line (ie every point on that line will map to somewhere else on that same line), so if lambda = 3 and corresponding eigenvector is_[2,1]then every point on the line y = 0.5x will map to three times itself on the same line.
If lambda =1 then each point will map to exactly itself, hence line of invariant points.

Reply 109

WillTheWiener

Original post by Qcomber

So in this case you could say an invariant line of the transformation would be y = 0.5x ?

Reply 110

B_9710

Original post by WillTheWiener

So in this case you could say an invariant line of the transformation would be y = 0.5x ?

Yes.

Reply 111

WillTheWiener

Original post by B_9710

Yes.

Thanks

Reply 112

B_9710

Original post by Qcomber

i can get that c=2t+5 and d=t.
but cant find a and b

Jordan-Gauss elimination is probably the best method here.

Reply 113

B_9710

Original post by Qcomber

i can get that c=2t+5 and d=t.
but cant find a and b

You'll have to introduce another parameter. The intersection of 3 4D planes is a 3D plane. Amazing right?
I got

\displaystyle \begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 5 \\ 0 \end{bmatrix} +\lambda \begin{bmatrix} -3 \\ 0 \\ 2 \\ 1 \end{bmatrix} +\mu \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}

Reply 114

fasdaman2212

FP4...is gonna destroy me

Reply 115

B_9710

Original post by 01binary

Are those equations always identical, I find that sometimes they aren't e.g. for lambda =1?
Well for part b) I found that the eigenvector for lamda =4 as (1 1 2) (x y z) respectively. The equation I got 3 times was x+y-z=0
Why would I need to split the vector (1, 1, 2) this is the part I am having trouble with.
Also is this in the AQA FP4 book anywhere

I was thinking, when you get the equation x+y-z=0 three times.
What is the equation of the plane you ask? (I know you didn't ask this, it's rhetorical) well it's x+y-z=0, in Cartesian form. I know that you have to find eigenvectors though of course. But it made me realise if you do the vector product of the 2 linearly independent vectors used to define the invariant plane then your answer should be some scalar multiple of the normal to the plane (i.e. the coefficients of the equation of the plane in Cartesian form).