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#1 Help Me ! » Matrix inverse proof » 2017-03-12 21:45:40

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Replies: 2

Let A be a n*n matrix and

Hi, I'm having trouble with the above question. I've made a start but I'm not sure if I'm approaching it correctly.
First of, there's a hint that suggests that I first consider the product AX. But if I do that, the dimensions of A and X are different so I would need to find the transpose of X?
so is it safe to assume

?

To find if A is invertible can I also do the following?
AX=XB (where B is some solution of A)


and for the formula:





#2 Re: Help Me ! » Lagrange multipliers » 2016-05-05 22:08:11

Thank you smile I was a little lost as what to do with those points. I'll have a go at it smile

#3 Re: Help Me ! » Lagrange multipliers » 2016-05-05 07:40:17

hi, I got the partial derivatives for f=x^2+y^2+z^2 (minimum distance from origin) and the given function, g=x^2-yz-6
so in terms of x: 2*x = 2*x*lambda (where lambda is the lagrange multiplier). I did the same for y and z.

#4 Help Me ! » Lagrange multipliers » 2016-05-05 07:19:31

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Replies: 5

Hi, I'm having trouble with a few sections of the question below:

9cOsHOL.jpg

For the first part, I found the minimum points to be (+- sqrt(6),0,0) .
Since this point is contained just on the x-axis, would it be safe to assume that if I take the normal from this point, it too will have just an x component and as a result pass through the origin?

For the section about intersection in the yz plane:
I used distance as one of my functions (y^2 +z^2) and the given surface, yz=-6. I found my minimum value to be (0,- sqrt(6),sqrt(6)) and (0,sqrt(6), - sqrt(6)) using lagrange.
But I'm unsure how I'd use the values I've found to explain why it's not a minima on the 3D surface.

Any help would be appreciated big_smile

#5 Re: Help Me ! » Lines and points » 2015-06-20 12:18:56

Hi bob,

I think I did step 1 correctly (?) and got: -s+7ts+5+t = 0, however having trouble with following step 2.

Just with how I did it to get the distance:
I put the two lines into the form
(1,1,2)+t(1,-3,2) and (0,1,2)+s(3,-2,-1)

then used, distance = (v(dot)w) /|w|

where v = v1- v2 = (1,1,2)-(0,1,2)
and w = d1 X d2 = (1,-3,2)X(3,-2,-1)
and subbed everything in to get the answer smile

#6 Help Me ! » Lines and points » 2015-06-19 23:54:48

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Replies: 4

Using the equations of two lines:
(1+t, 1-3t, 2+2t) and (3s, 1-2s, 2-s).

Find the distance between the two lines and find the uniquely determined points on the two lines that are this distance apart.

I found the distance to be 1/srt(3), having trouble with finding the two points though.

#8 Re: Help Me ! » L'hopital's Rule » 2015-06-18 23:38:52

Thank you smile
would you determine that its negative & positive infinity by looking at the graph it makes?

#9 Help Me ! » L'hopital's Rule » 2015-06-18 22:15:44

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Replies: 4

I applied L'hopital's  three times where i got:

where I thought the last was a determinant form and got the final value  = 0, but that's not correct. Could someone show me where i went wrong? smile

#10 Re: Help Me ! » Distance between point and line » 2015-06-03 20:27:29

Thanks for the help smile
wasn't careful with the signs

#11 Help Me ! » Distance between point and line » 2015-06-02 11:11:57

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Replies: 4

find the distance between the line r(t)= (1+t, 2-2t, -1+4t) and point (1,0,1)


With the above problem, I attempted it and thought I had a good idea of how to solve it, but the anwer I'm getting is different from the one that they've given.
Could someone help me find where i went wrong?

so, I calculated the distance between these two as
(1+t-1, 2-2t-0, -1+4t-1) = (t, 2-2t, 2+4t)

and then to find the value of t, from the equation of the line, i took d= (1, -2, 4)

and since these two are orthogonal, calculated the dot product of (t, 2-2t, 2+4t) and (1, -2, 4) and solved for t to get t=-4/21
and then from that length of t as (-4/21)^2 and squared it to get t=4/21

#12 Re: Help Me ! » Vectors » 2015-05-05 10:14:34

Why is it that u = u - v + v can be written as the length and absolute value?

#13 Help Me ! » Vectors » 2015-05-04 23:45:00

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Replies: 2

Prove that given any two vectors the following holds:
||u| - |v|| <= |u-v|
Hint start with u = u - v + v

I know that I'll need to use the triangular inequality at some point, however, not sure how to start it off.

#15 Help Me ! » L'Hopital's Rule » 2014-10-25 21:46:49

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Replies: 2

Hi;
Could someone explain how you would go about applying L'Hopital's Rule to the following limit?

I don't understand how they got some of the derivatives and how they were cancelled etc.

#16 Re: Help Me ! » Vector Help » 2014-08-26 13:04:50

Thank you Bob.
It's a long project. I think it's a lot clearer now, I'll post here again if I need more help.

#17 Help Me ! » Vector Help » 2014-08-26 06:01:32

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Replies: 4

Hi I'm having trouble with a part of this problem.
http://i.imgur.com/usIuMlB.png

I've been wanting to find  the points A and C.
I've calculated:
BD = D -B
     = (24.75, -10.4392, 0)
and the magnitude of BD = 26.87

but I'm completely  lost as where to go on from here.

#18 Science HQ » Circuit Analysis - Op Amps » 2014-08-22 21:10:17

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Replies: 0

Just wondering if anyone could help me with this problem?

What is the minimum value of RL if |Io| must be less than 75 mA for |VIN| < 0.8 V?
R2 = 8.5 kΩ, R1 = 9.9 kΩ

http://i.imgur.com/Os2mjp8.png

For my attempt I used the formulas:

(Vo/Vi) = -(R2/R1)

Io = Vo/(RL || (R2+R1)

and got an answer of 9.16 ohm, but that was incorrect. (It should be 19.8 ohms)

#20 Re: Help Me ! » Derivatives » 2014-06-04 14:32:47

Sorry about the delay; I think I've got it

?

Is the notation used okay?

#21 Re: Help Me ! » Derivatives » 2014-06-02 10:47:01

I might be misinterpreting this, so your saying

, then does that mean to get G'(x) you need to find the derivative of the stated inverse?

#22 Help Me ! » Derivatives » 2014-06-02 02:37:43

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Replies: 6

Need some help with this one.

#23 Help Me ! » Constant of Integration » 2014-05-23 13:42:43

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Replies: 2

Why do we need to add a constant, C, when we're solving an indefinite integral?

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