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leehuan's All-Levels-Of-Maths SOS thread (4 Viewers)

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leehuan

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What do you mean? Plug. in lambda = 2 to the matrix you posted. Then interpret the rows as equations as usual.
I mean, how would I know that the specific value was lambda = 2 and not something else like 5
 

InteGrand

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I mean, how would I know that the specific value was lambda = 2 and not something else like 5
Always try the ones that make diagonal elements 0. If a diagonal element is 0, it means the matrix is either not yet in row echelon form (e.g. Plugging lambda = 2 makes the matrix require further operations to get into row echelon form, since then the third column needs to be cleared up), or we get a zero row. In either case, you'll see eventually we have a zero row (when lambda = 2, we clear the third column and get a zero row in the third row), and when (and only when) that happens, we know there's potential for a lack of a solution.

If we just sub. in something random like 5, we see that the matrix ends up being in row echelon form with all rows leading, so there is a solution then. So in summary, try and only try values that'll make diagonal elements become 0, so that there will be a zero row, leading to a possibility of no solutions.
 
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leehuan

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Ah so to summarise, force out the diagonal matrix of coefficients to see what's going on?
 

InteGrand

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Ah so to summarise, force out the diagonal matrix of coefficients to see what's going on?
Yeah, if you're trying to find for which values of lambda the equations have no solution, just try and make diagonal entries 0, and investigate what happens (it's still possible that there solutions, because the right hand column may also be 0 where there's a zero row, depending on the situation with the lambdas, so this is why you need to investigate to see whether there are solutions or not).

(Basically try and only try values of lambda that'll immediately force a zero row, or will cause the matrix to NOT be in row echelon form.)
 

RealiseNothing

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quick question:
for 1/|x-1| > 1/|x+1|
I got the answer as x>0 and x≠1 or x≠-1
but the answer is just x>0 and x≠1
why can x=-1?
The textbook's answer is better though. You don't need to say x≠-1 since you already said x>0.
 

Drsoccerball

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How do you find the shortest distance in vectors.

1. How to derive it for a point to a line(in Cartesian form vs parametric form)
2. From line to line (Cartesian vs parametric)
 

leehuan

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Calling InteGrand and co.

I haven't done vectorial geometry like I said, to know how to do his question
 

InteGrand

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How do you find the shortest distance in vectors.

1. How to derive it for a point to a line(in Cartesian form vs parametric form)
2. From line to line (Cartesian vs parametric)
1) The method is described here in the 'vector formulation' section:

https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#A_vector_projection_proof



2) The method for finding the distance between two skew lines (non-parallel yet non-intersecting lines, which can exist only in dimension 3 or higher) in R^3 is described here:

https://en.wikipedia.org/wiki/Skew_lines#Distance .

If the lines are parallel, use the method from part 1). You can use any point on Line 1 as the external point to Line 2 and find its distance to Line 2, and this will be the answer.
 

parad0xica

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With these types of proofs, you choose at the end ^_^

Suppose the general formula: .

Then

.

Then you re-write the proof somehow making a magical guess of !
 
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Paradoxica

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Choose , then we have .

Now


as required.

A: double angle
B: s^2 + c^2 = 1
C: cosine attains max at 1
D: hypothesis
Note for part A and B, you should not have x/2. The substitution process is clear to me, and I have verified it myself, so it should definitely not be there.
 

Nailgun

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Choose , then we have .

Now


as required.

A: double angle
B: s^2 + c^2 = 1
C: cosine attains max at 1
D: hypothesis
wait you actually can do maths

i thought you were a troll account

wat
 
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