Linear algebra help!

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hayood

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Feb 16, 2010
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Show that there exists scalars c1, c2, and c3 , not all 0, such that c1(?2, 9, 6) + c2(?3, 2, 1) + c3(1, 7, 5) = (0, 0, 0) (the zero vector) (The vectors (?2, 9, 6), (?3, 2, 1) and (1, 7, 5) are called linearly dependent. )

my answer:
i used reduced row echelon to try and get a free variable but it never works..please help:)
 
hayood said:
Show that there exists scalars c1, c2, and c3 , not all 0, such that c1(?2, 9, 6) + c2(?3, 2, 1) + c3(1, 7, 5) = (0, 0, 0) (the zero vector) (The vectors (?2, 9, 6), (?3, 2, 1) and (1, 7, 5) are called linearly dependent. )

my answer:
i used reduced row echelon to try and get a free variable but it never works..please help:)
Click to expand...

You are following the right procedure. After reducing you should get
[1 0 1
0 1 -1
0 0 0]

then you assign \(\displaystyle c_2\) a variable \(\displaystyle c_2=x\) and you substitute it into the other equations
to get \(\displaystyle c_1\) and \(\displaystyle c_2\) in terms of \(\displaystyle c_3\)

then you get a column vector with x and you can pick any value for x. Normally people us 1.

it that clear?
 
i don't get why c2 =x shouldn't it be c1 +c3=0, c2 -c3 =0 , and 0 =0 which one is the free variable?
if i choose c1 =1 what do i choose for c2?

thanks:)
 
hayood said:
i don't get why c2 =x shouldn't it be c1 +c3=0, c2 -c3 =0 , and 0 =0 which one is the free variable?
if i choose c1 =1 what do i choose for c2?

thanks:)
Click to expand...


When you assign \(\displaystyle c_3\) as x....you are making it a variable that you can give any value you want. When you put that into your equations you then have
\(\displaystyle c_1\) and \(\displaystyle c_2\) in terms of your variable. all of your c's are related to each other by x. if you give x a value you know what
\(\displaystyle c_1\), \(\displaystyle c_2\) and \(\displaystyle c_3\) are.

is that okay with you?
 
so i just say that let c3 =x and c1 =1 and then substitute that in the equations to get c2 ?
sorry i still find it unclear.
 
no you don't make \(\displaystyle c_1=1\) you make x equal to 1 and then you get \(\displaystyle c_3=1\) \(\displaystyle c_2=1\) and \(\displaystyle c_1=-1\) because of your equations...
\(\displaystyle c_1 +c_3=0\) and \(\displaystyle c_2 -c_3 =0\) Plug in 1 for \(\displaystyle c_3\) and that is what you get.


you can check it. Plug \(\displaystyle c_3=1\) \(\displaystyle c_2=1\) and \(\displaystyle c_1=-1\) into the original problem. it will work out.
 
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