Let v1 = -4 -1
-2

v2 = -3
1
-2

v3= 1
-5
2

and H = Span{v1,v2,v3} . Note that v3 = 2v1 - 3v2.

Which of the following sets form a basis for the subspace H, i. e., which sets form an efficient spanning set containing no unnecessary vectors?

a. {V1, V2, V3}
b. {V1, V2}
c. {V1,V3}
d. {V2,V3}

We're told (and we can confirm) that , so is a linear combination of the other two vectors.

This means H is sufficiently spanned by ; no need for the third vector.

But this also means we can write either as a linear combination of , and as a lin. com. of . So any set of these three vectors taken two at a time will span the subspace H. Hence all of b, c, and d are acceptable.

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x

a

step-by-step explanation: