Quick question on linear dependence

[98.40_for_sure]

If they ask you to prove if 3 vectors are linearly dependent, is this right? or is it the other way around

Proof by simultaneous equations, where linearly dependent is mu + nv + pw = 0, where m = n = p = 0?
Or is it when they DONT equal 0?

[TrueTears]

i think thats when they are linearly independent.

um to prove linear dependence you must show a = kb+qc for some vectors a, b, c, i won't explain the intuition behind this as you probably know why, this is probs the most simplest method...

[fady_22]

If they ask you to prove if 3 vectors are linearly dependent, is this right? or is it the other way around

Proof by simultaneous equations, where linearly dependent is mu + nv + pw = 0, where m = n = p = 0?
Or is it when they DONT equal 0?

When they don't equal zero.
Also, don't introduce three variables, you'd be making it difficult for yourself.
Do this instead:
w=mu+pw

[98.40_for_sure]

Ohh wait i misread the solution soz,

but the neap 2010 spesh exam 1 solution uses 3 variables? how do you do it with 2?

see attachment

[SixWinged]

m, n and p cannot equal 0. If you think about it, setting all these to 0 would allow any set of vectors to be expressed as 0 = 0 + 0.

In my opinion, the easiest way to prove linear dependence is to us (constant)a + (constant2)b = c, that way there's only two variables to solve for, plus using the (constant)a + (constant2)b + (constant3)c = 0 method quite often just leaves you with two of the constants in terms of the third anyway, easier to just let constant3 = 1.

[fady_22]

So, as I said:
w=mu+nv
Collect and equate coefficients:
m+2n=1 ...1
m+n=2 ...2
1=-m-2n ...3

Then I would solve 1 and 2 separately, and then substitute the values found for m and n into 3, and show that these are not solutions for 3.
As there is no common solution, the set of vectors are linearly independent.

(You can see from 1 and 3 that these are two parallel lines, hence there will be no solutions anyway. But for a question with many marks, its best to show all working).

[98.40_for_sure]

Ohh yeah i forgot about solving from (1) and (2) then subbing into (3) to prove. thanks fady

[TrueTears]

99forshiz dont do it the companies way, i've seen another paper do it that way in a trial exam for a student yesterday, it's stupid and takes more work