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 $\mathbb R^3$.