Linear Algebra - Linear Combination of Vectors

Question

Consider the vectors u = (1, 2,-1) and w = (6, 4, 2) in R3, show that w = (9, 2, 7) is

a linear combination of u and v and that w0 = (4, -1, 8) is not a linear combination

of u and v.

Answer 1

so we will start by making a matrix

[1 6 | 9]

[2 4 | 2]

[-1 2 | 7]

now we start row reducing R2=R2-2R1 R3=R3+R1

[1 6 | 9]

[0 -8 | -16]

[0 8 | 16]

Now do R3=R3+R2 and R2=-1/8R2

[1 6 | 9]

[0 1 | 2]

[0 0 | 0]

Now do R1=R1-6R2

[1 0 | -3]

[0 1 | 2]

[0 0 | 0]

so w=-3u+2v

Now do the same thing for w0 (skipping to where R2 is already reduced since same steps):

[1 6 | 9]

[0 1 | 2]

[4 -1 | 8]

so now R3=R3-4R1

[1 6 | 9]

[0 1 | 2]

[0 -25 | -28]

now R3=R3+25R2

[1 6 | 9]

[0 1 | 2]

[0 0 | 22]

by the bottom row the vector w0 is not a linear combination of u and v

Answer 2

As v is not a multiple of u and neither is the 0 vector, we calculate the determinant. (Note that the second vector is mislabeled and should be v.) If it is 0, the third vector is a linear combination of the first two. If the determinant is not 0, it is not a linear combination.

For

1 2 -1

6 4 2

9 2 7

the determinant is

1(4*7-2*2) -2(6*7-2*9) -1(6*2-4*9) =

1(24)-2(24)-1(-24) =

24 - 48 - -24 = 0

so w is a linear combination of u and v

1 2 -1

6 4 2

4 -1 8

has determinant

1(4*8-2*-1) - 2(6*8-2*4) -1(6*-1-4*4) =

1(34) -2(40) -1(-22) =

34 - 80 + 22 =

-24

so w0 is not a linear combination of u and v