Assume that the set {vector x, vector y, vector z} is linearly independent.

Prove that the set

{5(vector x) + 2(vector y) + vector z,-2(vector x)-6(vector y), 4(vector x) + 3(vector y) + 8(vector z)}

is linearly independent.

### 1 Answer | Add Yours

`Let```

`u=5x+2y+z `

`v=-2x-6y`

`w=4x+3y+8z`

`be the vectors which are linearly independent.`

`alphau+betav+gamma w=0`

`Then`

`alpha=beta=gamma=0`

`alpha(5,2,1)+beta(-2,-6)+gamma(4,3,8)=(0,0,0)`

`5alpha-2beta+4gamma=0`

`2alpha-6beta+3gamma=0`

`alpha+8gamma=0`

`[[5,-2,4],[2,-6,4],[1,0,8]]` the cofficient matrix has rank 3.

So above homogeneous system of equation has only solution

`alpha=beta=gamma=0`

Thus our asumption is true and vector u,v, and w are linearly independent.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes