Question

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 is linearly dependent and does not span R3. Sis linearly independent and spans R3. Sis linearly independent and does not span R3.

Answer 1

`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`.