# Let v1=(5,2,0), v2=(3,4,1), v3=(1,3,2), v4=(0,1,1) Is{v1,v2,v3,v4} a generating system for R^3. Is is a basis for R^3? why/why not?

If is generating `RR^3` iff any vector (x,y,z) of` RR^3` can be written in the form `av_1+bv_2+cv+3+dv+4=(x,y,z)` for some (a,b,c,d) in `RR^4.`

Let's solve the condition for (a,b,c,d)

`(x,y,z)=(5a,2a,0)+(3b,4b,b)+(c,3c,2c)+(0,d,d)`

`x=5a+3b+c`

`y=2a+4b+3c+d`

`z=b+2c+d`

Let's try to find a solution with b=0

Then

`x=5a+c`

`z=2c+d`

`c=x-5a`

`d=z-2c=z-2x+10a`

y=2a+3(x-5a)+z-2x+10a=12a+3x-15a-2x=-3a+x+z

Therefore

`c=x-5a`

`d=z-2x+10a`

`3a=x+z-y `

`a=(x+z-y)/3`

`c=x-5a=x-5(x+z-y)/3`

`d=z-2x+10(x+z-y)/3`

Remember b=0

Therefore for any vector (x,y,z) we found a linear combination such that `(x,y,z)=av_1+bv_2+cv_3+dv_5. ` The expression of a,b=0,c, d are given above.** Therefore the 4 vectors are generating** `RR^3.`

`RR^3` is a 3 dimensional vector space. Any basis of `RR^3 ` contains 3 vectors. **Therefore the 4 vectors** `v_1, v_2, v_3, v_4` **can't be a basis.**