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Anastasy [175]
3 years ago
10

Explain why S = {(5, 4, -2), (-15, -12, 6), (10, 8, -4)} is NOT a basis for R^3 (1 point Sis linearly dependent and spans R3 S i

s linearly dependent and does not span R3. Sis linearly independent and spans R3. Sis linearly independent and does not span R3.
Mathematics
1 answer:
earnstyle [38]3 years ago
7 0

S would be a basis for \mathbb R^3 if

(1) the vectors in S are independent, and

(2) the vectors span \mathbb R^3.

  • Linear independence requires that c_1=c_2=c_3=0 is the only solution to

c_1(5,4,-2)+c_2(-15,-12,6)+c_3(10,8,-4)=(0,0,0)

These vectors are not linearly independent because if c_1=3, c_2=1, and c_3=0, we have

3(5,4,-2)+(-15,-12,6)=(15-15,12-12,-6+6)=(0,0,0)

so S is not a basis for \mathbb R^3.

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