Find homogeneous system of linear equations whose solution space is: V = span((1,-2,4,3),(1,-1,6,4),(3,-8,8,3)).First I found vectors were linearly dependent, so I discarded the third vector to form a new base. Next I figured system will have 2 free variables and if we would use gauss-jordan elimination on matrix of a system we would get this: [ 1 0 ∗ ∗ 0 1 ∗ ∗ ]

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Find homogeneous system of linear equations whose solution space is: V = span((1,-2,4,3),(1,-1,6,4),(3,-8,8,3)).First I found vectors were linearly dependent, so I discarded the third vector to form a new base. Next I figured system will have 2 free variables and if we would use gauss-jordan elimination on matrix of a system we would get this:      [                            1                          0                          ∗                          ∗                                      0                          1                          ∗                          ∗                      ]

Rachael Conner · Accepted Answer

Actually the 3 vectors are linearly independent. For example, it is not hard to check that      |                            1                          −          2                          4                          3                                      1                          −          1                          6                          4                                      3                          −          8                          8                          3                                      1                          0                          0                          0                      |    =  −  32  ≠  0So we expect there will be just one equation. Suppose it is that the vector (  (  w  ,  x  ,  y  ,  z  )) must satisfy   a  w  +  b  x  +  c  y  +  d  z  =  0. Then we know that   a  ,  b  ,  c  ,  d must satisfy:(1)   a  −  2  b  +  4  c  +  3  d  =  0, (2)   a  −  b  +  6  c  +  4  d  =  0, (3)   3  a  −  8  b  +  8  c  +  3  d  =  0Taking (2)-(1), we get   b  +  2  c  +  d  =  0, and taking 2(2)-(1), we get   a  +  8  c  +  5  d  =  0. Substituting in (3) gives   d  =  0. So the only possible equation is essentially   8  w  +  2  x  −  y  =  0. It is easy to check that the three given points all satisfy it and hence all linear combinations of those points will also satisfy it.

Find homogeneous system of linear equations whose solution space is: V = span((1,-2,4,3),(1,-1,6,4),(3,-8,8,3)).

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