# prove that the vectors (1,2,1),(2,1,0)(1,-1,2) form a basis over R^3

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You need to form a system of 3 homogeneous equations in `barv_1, barv_2,barv_3` , such that:

`barv_1(1,2,1) + barv_2(2,1,0) + barv_3(1,-1,2) = (0,0,0)`

`(barv_1+2barv_2+barv_3,2barv_1+barv_2-barv_3,barv_1+2barv_3) = (0,0,0)`

`{(barv_1+2barv_2+barv_3 = 0),(2barv_1+barv_2-barv_3 = 0),(barv_1+2barv_3 = 0):}`

The system of homogeneous equations has a non-zero solution if `Delta = 0` , hence, you need to evaluate the determinant of matrix of coefficients of the system, such that:

`Delta = [(1,2,1),(2,1,-1),(1,2,0)]` = `0 + 4 - 2 - 1 + 2 - 0 = 3!= 0`

Since `Delta != 0` , hencem the system has only the trivial solution `barv_1 = barv_2 = barv_3 = 0` .

The vectors `barv_1,barv_2, barv_3` also generate `R^3` , such that:

`barv_1(1,2,1) + barv_2(2,1,0) + barv_3(1,-1,2) = (x,y,z)`

`(x,y,z) in R^3` and x,y,z are not simultaneously 0

`{(barv_1+2barv_2+barv_3 = x),(2barv_1+barv_2-barv_3 = y),(barv_1+2barv_3 = z):}`

Since it was shown that `Delta!=0` , hence, the solution of the system is not trivial solution.

**Considering all the above yields, by definition, that the given vectors `(1,2,1),(2,1,0),(1,-1,2)` form a basis over the vector space `R^3` .**

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