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leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

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Paradoxica

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That's what I was using lel

btw was the triangle area correct?

it almost cancels out perfectly to 4/3 so i think ive made a mistake somewhere
Pretty sure the area of the triangle has 4, not 8.

This can be observed by assuming the result desired is true.
 

leehuan

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I do not compute. How did they derive this?
 

leehuan

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The question is obviously trivial





Obviously I can tell how they got their answer. But would there be any reason as to mine being incorrect here and do I have to give my answer the same way they did?
 

InteGrand

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The question is obviously trivial





Obviously I can tell how they got their answer. But would there be any reason as to mine being incorrect here and do I have to give my answer the same way they did?
 
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leehuan

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Oh of course. Thanks.

(This is why nobody should do maths whilst having a headache)
 

leehuan

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The entire question:


Given answers:


Now, my working thus far has everything besides lambda=-2 at this point.





 

seanieg89

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When lambda=-2, the bottom two rows will give you contradictory expressions for z.

Not sure if you have learned about determinants yet, but they give you a v. quick way of determining when a matrix is invertible (ie when Ax=y has a unique solution). This reduces such problems to checking the isolated values of lambda that make A singular to check whether the resulting systems have no solutions or infinitely many.
 

Drongoski

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The question is obviously trivial





Obviously I can tell how they got their answer. But would there be any reason as to mine being incorrect here and do I have to give my answer the same way they did?
To add to the discussion:

The 2 equations, i.e. the 2 planes are actually the 1 plane as the 2 planes are identical. So the solution set consists of all the points that lie on the single plane. The given answer is the usual way of specifying this set.
 

leehuan

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When lambda=-2, the bottom two rows will give you contradictory expressions for z.

Not sure if you have learned about determinants yet, but they give you a v. quick way of determining when a matrix is invertible (ie when Ax=y has a unique solution). This reduces such problems to checking the isolated values of lambda that make A singular to check whether the resulting systems have no solutions or infinitely many.
Yea problem, we haven't. So I wouldn't have been able to spot that lambda=-2 gives two contradictions just by looking at the matrix
 

InteGrand

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Yea problem, we haven't. So I wouldn't have been able to spot that lambda=-2 gives two contradictions just by looking at the matrix
Don't need to know about determinants for this. Sub. lambda = -2, then equation 2 (row 2) says that -z = 1, whilst row 3 says that -12z = 3, a contradiction.
 

leehuan

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Don't need to know about determinants for this. Sub. lambda = -2, then equation 2 (row 2) says that -z = 1, whilst row 3 says that -12z = 3, a contradiction.
Yea but I wouldn't have seen that just by staring at my matrix
 

Green Yoda

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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?
 

InteGrand

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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?
Because then the denominator of the RHS would be 0, and we can't divide by 0.

Edit: oh you said why *can*. Well, it can't be -1 in fact, as you had.
 
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