c1v1+c2v2+c3v3+c4v4=0
c1(1,0,0)+c2(1,1,0)+c3(3,1,0)+c4(0,−2,0)=0
From the above equation we get the following equations:
c1+c2+3c3=0
c2+c3−2c4=0
Simplifying we get:
c1=2c3−2c4
c2=2c4−c3
The {v1,v2 } spans span of subspace and are linearly independent.
Indeed,
c1v1+c2v2=0, c1(1,0,0)+c2(1,1,0)=0.
We obtain the following equations
c1+c2=0,
c2=0.
Hence c2=0,c1=0.
Thus, {v1,v2 } is the basis.
Checking option (A)
v3=2v1+v2, v4=2v1−2v2.
Thus, v1 and v2 form a basis for span {v1,v2,v3,v4}.
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