Find a basis for the solution set of the given homogeneous linear system 3 x 1 + x 2 + x 3 = 0 6 x 1 + 2 x 2 + 2 x 3 = 0 − 9 x 1 − 3 x 2 − 3 x 3 = 0I do what I know I need to do. First I get the solution set of the system by reducing like this: ( 3 1 1 6 2 2 − 9 − 3 − 3 ) ⇝ ( 3 1 1 0 0 0 0 0 0 ) ⇝ ( 1 1 / 3 1 / 3 0 0 0 0 0 0 ) So I know x → = [ x 1 x 2 x 3 ] = [ 1 − 1 3 r − 1 3 s r s ] That being the general solution.Now, giving the values for r and s according to the standard vectors i, j x → = [ x 1 x 2 x 3 ] = [ 1 − 1 3 r − 1 3 s r s ] = r [ 2 3 1 0 ] + s [ 2 3 0 1 ] From my results, the basis will be: ( [ 2 3 1 0 ] , [ 2 3 0 1 ] )But instead, the book answer ( [ − 1 3 0 ] , [ − 1 0 3 ] )Any idea on what I'm doing wrong?

Question

Find a basis for the solution set of the given homogeneous linear system  3      x    1    +      x    2    +      x    3    =  0  6      x    1    +  2      x    2    +  2      x    3    =  0  −  9      x    1    −  3      x    2    −  3      x    3    =  0I do what I know I need to do. First I get the solution set of the system by reducing like this:      (                            3                          1                          1                                      6                          2                          2                                      −          9                          −          3                          −          3                      )    ⇝      (                            3                          1                          1                                      0                          0                          0                                      0                          0                          0                      )    ⇝      (                            1                          1                  /          3                          1                  /          3                                      0                          0                          0                                      0                          0                          0                      )  So I know       x    →    =      [                                        x            1                                                            x            2                                                            x            3                                ]    =      [                            1          −                      1            3                    r          −                      1            3                    s                                      r                                      s                      ]  That being the general solution.Now, giving the values for   r and   s according to the standard vectors   i,   j      x    →    =      [                                        x            1                                                            x            2                                                            x            3                                ]    =      [                            1          −                      1            3                    r          −                      1            3                    s                                      r                                      s                      ]    =  r      [                                        2            3                                                1                                      0                      ]    +  s      [                                        2            3                                                0                                      1                      ]  From my results, the basis will be:  (      [                                        2            3                                                1                                      0                      ]    ,      [                                        2            3                                                0                                      1                      ]    )But instead, the book answer  (      [                            −          1                                      3                                      0                      ]    ,      [                            −          1                                      0                                      3                      ]    )Any idea on what I&#039;m doing wrong?

laure6237ma · Accepted Answer

Check your expression for       x    →  . It should be      x    →    =      [                                        x            1                                                            x            2                                                            x            3                                ]    =      [                            −                      1            3                    r          −                      1            3                    s                                      r                                      s                      ]    =  r      [                            −                      1            3                                                1                                      0                      ]    +  s      [                            −                      1            3                                                0                                      1                      ]

Layla Velazquez · Accepted Answer

don&#039;t follow this step.      x    →    =      [                                        x            1                                                            x            2                                                            x            3                                ]    =      [                            1          −                      1            3                    r          −                      1            3                    s                                      r                                      s                      ]    =  r      [                                        2            3                                                1                                      0                      ]    +  s      [                                        2            3                                                0                                      1                      ]  Wouldn&#039;t you want to write this as:      x    →    =      [                                        x            1                                                            x            2                                                            x            3                                ]    =      [                            1          −                      1            3                    r          −                      1            3                    s                                      r                                      s                      ]    =  r      [                                                      −              1                        3                                                1                                      0                      ]    +  s      [                                                      −              1                        3                                                0                                      1                      ]  If you write out the eigenvalues, you get   λ  =  1  ,  0  ,  0The corresponding eigenvectors are:  (  1  ,  0  ,  0  )  ,  (  −  1  /  3  ,  0  ,  1  )  ,  (  −  1  /  3  ,  1  ,  0  ) and these should look familiar to your solution.

Find a basis for the solution set of the given homogeneous linear system 3 x 1 </ms

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