Recent content by Cmunro

Aha, got it, thank you!

Hi, I am just starting out in calculus, and I'm not sure how to work through this type of question: f' \frac{3x-6}{x} So I have: (3x-6)(x^{-1}) then (3x -6)(-x^{-2}) Now what? (3)(-x^{-2})? The book gives an answer of 6(x^{-2}) Thanks in advance

ahh yes - I see what you mean! It is added to my answer - I always forget to give the parameters. Anyway, I appreciate all the time you have given!

Oh sorry! I moved from variations of X to a,b,c,d as it became much easier to manage that way: i.e matrix x: [a b] [c d] Where the equations become four as before, but cancel out to give the above simplifications of a=b, b=-c, b=-d. So x is basically: [a a] [-a -a] It is essentially the...

I shot off an email to my teacher, and he said that Gauss-Jordan is a much more complicated way to work it out. The hint is that there are an infinite number of solutions. So I have canceled down the equations to look like this: a=b b=-c b=-d It seems very simple.. so I'm not sure...

I'm really stumped.. because I can't see how to get any other number than zero on the right side.

I'm completely at a loss for how to apply this to the question at hand, and I don't believe I have been taught this. I understand the system you have described, I have no numbers at all to work with though for my question...right?

No I'm not. A quick search of my textbook (Mathematics for the international student IB SL..John Owen et. al -if that means anything) did not give any results. I searched online. Would you say this gives a good explanation? http://en.wikipedia.org/wiki/Gauss-Jordan_elimination if so, I'll...

Ok so.. 4 equations: X1=2X1 -X2 (1) X2= -X1 +2X2 (2) X1+2X3 =2X3-X4 (3) X2+2X4=-X3+X4 (4) Sub (1) into (2) 2X1-X2=-X1+2X2 and X2=X1 Also, manipulation of (3) gives X2 (likewise X1)= -X4 Sub this info into (4) -X4 +2X4 = -X3 +X4 which gives: X4=-X3/2 but now I'm stuck. Seems to become...

Oh dear. Brain fart. For some reason I just took the top bit and the bottom bit as two equations.

Ok.. I'm sure this must be very obvious - but I'm just not sure where I am going with this. so: [1 0][X1 X2]...[X1 X2][2 -1] [1 2][X3 X4] =[X3 X4][-1 2] [(X1)...(X2)]...[(2X1-X2)...(-X1+2X20] [(X1+2X3) (X2+2X4)] =[(2X3-X4)...(-X3+X4)] Again, sorry, bad formatting. Ignore the...

"This is a good piece of advice, why don't you listen to it?" Oh I thought he that by linear equations he meant just that. Right, now I see what you mean. I'll have a go at it that way and let you know how I do.

Sorry I have been away and did not have internet access. Thank you all so much! (I'm sorry for the delay in thanking you) It all gets very complicated with the pluses and minuses everywhere, and I find myself losing numbers here and there which is hardly useful. Anyway, thank you, now I see...

I tried.. at least.. If I moved the A around it would become X= A(inv)XB likewise the B: AXB(inv)=X or.. AX=XB A=XBX(inv) Then it just seems to be circular. I just wonder if there is a rule I have overlooked. As I know that AB does not equal BA..

The question is as follows: Find X: [1 0]...[2 -1] [1 2] X= X [-1 2] (sorry that is the only way I thought to show the matrix..ignore the dots) My attempt: I tried to work this out by making it simpler: AX=XB But even if I inverse the A or the B.. I still can't get X on its own. Any...