I am given the following matrix A and I need to find a nullspace of this matrix. A = ( 2 1 4 − 1 1 1 1 1 1 0 3 − 2 − 3 − 2 − 5 0 ) I have found a row reduced form of this matrix, which is: A ′ = ( 1 0 3 − 2 0 1 − 2 3 0 0 0 0 0 0 0 0 ) And then I used the formula A ′ x = 0, which gave me: A ′ x = ( 1 0 3 − 2 0 1 − 2 3 0 0 0 0 0 0 0 0 ) ( x 1 x 2 x 3 x 4 ) = ( 0 0 0 0 ) Hence I obtained the following system of linear equations: { x 1 + 3 x 3 − 2 x 4 = 0 x 2 − 2 x 3 + 3 x 4 = 0 So I just said that x 3 = α, x 4 = β and the nullspace is: n u l l s p a c e ( A ) = { 2 β − 3 α , 2 α − 3 β , α , β ) | α , β ∈ R }Is my thinking correct?

Question

I am given the following matrix A and I need to find a nullspace of this matrix.  A  =      (                            2                          1                          4                          −          1                                      1                          1                          1                          1                                      1                          0                          3                          −          2                                      −          3                          −          2                          −          5                          0                      )  I have found a row reduced form of this matrix, which is:      A    ′    =      (                            1                          0                          3                          −          2                                      0                          1                          −          2                          3                                      0                          0                          0                          0                                      0                          0                          0                          0                      )  And then I used the formula       A    ′    x  =  0, which gave me:      A    ′    x  =      (                            1                          0                          3                          −          2                                      0                          1                          −          2                          3                                      0                          0                          0                          0                                      0                          0                          0                          0                      )        (                                        x            1                                                            x            2                                                            x            3                                                            x            4                                )    =      (                            0                                      0                                      0                                      0                      )  Hence I obtained the following system of linear equations:      {                                        x            1                    +          3                      x            3                    −          2                      x            4                    =          0                                                  x            2                    −          2                      x            3                    +          3                      x            4                    =          0                        So I just said that       x    3    =  α,       x    4    =  β and the nullspace is:  n  u  l  l  s  p  a  c  e  (  A  )  =  {  2  β  −  3  α  ,  2  α  −  3  β  ,  α  ,  β  )         |       α  ,  β  ∈      R    }Is my thinking correct?

Zayden Wiley · Accepted Answer

Since       x    1    =  2      x    4    −  3      x    3   and       x    2    =  2      x    3    −  3      x    4    ⇒if   (      x    1    ,      x    2    ,      x    3    ,      x    4    )  ∈ nullspace(A):  (      x    1    ,      x    2    ,      x    3    ,      x    4    )  =  (  2      x    4    −  3      x    3    ,  2      x    3    −  3      x    4    ,      x    3    ,      x    4    )  =      x    3    (  −  3  ,  2  ,  1  ,  0  )  +      x    4    (  2  ,  −  3  ,  0  ,  1  )So Nullspace   (  A  )  =  ⟨  (  −  3  ,  2  ,  1  ,  0  )  ,  (  2  ,  −  3  ,  0  ,  1  )  ⟩.