Find a Basis for the solution set

[digitol87]

3x1 + x2 + x3 = 0
6x1 + 2x2 + 2x3 = 0
-9x1 - 3x2 - 3x3 = 0

I'm not sure how to approach this problem. I've rewritten these equations as a matrix

[3 1 1]
[6 2 2]
[-9 -3 -3]

Reduced Echelon from gave me this
[3 1 1]
[0 0 0]
[0 0 0]

Am I approaching this the wrong way?
What should I do next? Please help.
Thank You.

 

[Mark44]

3x1 + x2 + x3 = 0
6x1 + 2x2 + 2x3 = 0
-9x1 - 3x2 - 3x3 = 0

By inspection it can be seen that the 2nd equation is 2 times the first, and the 3rd is -3 times the first. In essence, you have the same equation written three times.

I'm not sure how to approach this problem. I've rewritten these equations as a matrix

[3 1 1]
[6 2 2]
[-9 -3 -3]

Reduced Echelon from gave me this
[3 1 1]
[0 0 0]
[0 0 0]

Am I approaching this the wrong way?
What should I do next? Please help.
Thank You.

Solve the first equation to get
x₁ = -(1/3)x₂ - (1/3)x₃
x₂ = x₂ + 0x₃
x₃ = 0x₂ + x₃

The 2nd and 3rd equations above are obviously true.

If you stare at this system awhile, you might see that any vector <x₁, x₂, x₃> in this set can be written as a linear combination of two vectors that happen to be linearly independent.