Show that S 1 1 2 3 2 4 0 1 0is a basis for R3SolutionDenot

Show that S = {[1 1 -2], [3 2 -4], [0 1 0]}is a basis for R^3

Denote transpose by: \'

ie A^T=A\'

So, S={v1,v2,v3}

v1=(1,1,-2)\',v2=(3,2,-4)\',v3=(0,1,0)\'

e1=v2-2v1=(1,0,0)\'

e1+v3-v1=(0,0,2)\'

Hence, vectors in S spans all the standard basis vectors of R3 and hence S is a basis of R3

$Show that S = {[1 1 -2], [3 2 -4], [0 1 0]}is a basis for R^3SolutionDenote transpose by: \' ie A^T=A\' So, S={v1,v2,v3} v1=(1,1,-2)\',v2=(3,2,-4)\',v3=(0,1,0)$

Show that S 1 1 2 3 2 4 0 1 0is a basis for R3SolutionDenot

Solution

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