Find a basis for the set of all vectors of the form (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c).

This made so much sense in the book.

Find a basis for the set of all vectors of the form (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c).

This made so much sense in the book.

This made so much sense in the book.

So the basis is just ? It's that easy?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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So the basis is just ? It's that easy?

No...that is not the answer

Could you give me another hint, please?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
**Contact:**

therefore the vectors are not linearly independent

So this is where I have to do the vector-sum equals zero thing? When I solve the matrix, I get...

So a=-c, b=2c. If c=-1, then a=1,b=-2 and (1,2,-1,3)+(4,-10,8,-2)+(-5,8,-7,-1) = (0,0,0,0). So the answer is just the first two vectors, (1,2,-1,3),(-2,5,-4,1)?

So a=-c, b=2c. If c=-1, then a=1,b=-2 and (1,2,-1,3)+(4,-10,8,-2)+(-5,8,-7,-1) = (0,0,0,0). So the answer is just the first two vectors, (1,2,-1,3),(-2,5,-4,1)?

- Martingale
**Posts:**333**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
**Contact:**

So this is where I have to do the vector-sum equals zero thing? When I solve the matrix, I get...

So a=-c, b=2c. If c=-1, then a=1,b=-2 and (1,2,-1,3)+(4,-10,8,-2)+(-5,8,-7,-1) = (0,0,0,0). So the answer is just the first two vectors, (1,2,-1,3),(-2,5,-4,1)?

those two vectors will work

I don't know why I'm having so much trouble with this. Thank you for your help!