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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.

Respuesta :

LammettHash

[tex]S[/tex] would be a basis for [tex]\mathbb R^3[/tex] if

(1) the vectors in [tex]S[/tex] are independent, and

(2) the vectors span [tex]\mathbb R^3[/tex].

  • Linear independence requires that [tex]c_1=c_2=c_3=0[/tex] is the only solution to

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

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

[tex]3(5,4,-2)+(-15,-12,6)=(15-15,12-12,-6+6)=(0,0,0)[/tex]

so [tex]S[/tex] is not a basis for [tex]\mathbb R^3[/tex].

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