Find a Basis for the solution set

  • Thread starter digitol87
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
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7,520
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.
 

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