Determine a basis for the solution set of the homogeneous system x 1 + x 2 + x 3 = 0 3 x 1 + 3 x 2 + x 3 = 0 4 x 1 + 4 x 2 + 2 x 3 = 0 Then the augmented matrix is:Reduced Row Echelon Form → [ 1 1 0 0 0 0 1 0 0 0 0 0 ]

Question

Determine a basis for the solution set of the homogeneous system                              x          1                +                  x          2                +                  x          3                                    =        0                            3                  x          1                +        3                  x          2                +                  x          3                                    =        0                            4                  x          1                +        4                  x          2                +        2                  x          3                                    =        0            Then the augmented matrix is:Reduced Row Echelon Form   →      [                            1                          1                          0                          0                                      0                          0                          1                          0                                      0                          0                          0                          0                      ]

Kaydence Washington · Accepted Answer

The first equation in the Reduced Row Echelon Form tells you that we need       x    1    +      x    2    =  0 and the second equation says       x    3    =  0. So If we take       x    2    =  t for   t  ∈  R then we must have       x    1    =  −      x    2    =  −  t and       x    3    =  0.Thus we have a 1-dimensional solution space determined by the vector  (      x    1    ,      x    2    ,      x    3    )  =  (  −  1  ,  1  ,  0  )This follows from the above discussion since taking       x    2    =  t corresponds to scalar multiplication by t on   (  −  1  ,  1  ,  0  ).

seupeljewj · Accepted Answer

The equations to be satisfied are       x    1    +      x    2    +      x    3    =  0   3      x    1    +  3      x    2    +      x    3    =  0   4      x    1    +  4      x    2    +      x    3    =  0The first thing I notice is that if we multiply the first equation by 3 and subtract the second equation we have   2      x    3    =  0 so any solution to this system of equations must have       x    3    =  0. Setting x3=0 in these equations, we have       x    1    +      x    2    =  0,   3      x    1    +  3      x    2    =  0,   4      x    1    +  4      x    2    =  0All of which are equivalent to       x    1    +      x    2    =  0 or       x    2    =  −      x    1  . Any solution to this system of equations is for the form   (      x    1    ,      x    2    ,      x    3    )  =  (      x    1    ,  −      x    1    ,  0  )  =      x    1    (  1  ,  −  1  ,  0  ).That is a one dimensional subspace of       R    3   with basis   {  (  1  ,  −  1  ,  0  )  }. (Of course,   {  (  −  1  ,  1  ,  0  )  } is equivalent.)

Determine a basis for the solution set of the homogeneous system <mtable columnalign="right lef

Answered question

Answer & Explanation

New Questions in Algebra I