Show that the set of vectors {(−4,1,3),(5,1,6),(6,0,2)} does not span R 3 , but that it does span the subspace of R 3 consisting of all vectors lying in the plane with equation x + 13y − 3z = 0.

The vectors does not span R3 and only span a subspace of R3 which satisfies x+13y-3z=0

Explanation:

The vectors are given as

Now if the vectors  would span the , the rank of  the consolidated matrix will be 3 if it is not 3 this indicates that the vectors does not span the .

So the matrix is given as

In order to calculate the rank, the matrix is reduced to the Row Echelon form as

As the Rank is given as number of non-zero rows in the Row echelon form which are 2 so the rank is 2.

Thus this indicates that the vectors does not span

Now for any vector the corresponding equation is formulated by using the combined matrix which is given as  for any arbitrary vector and the coordinate as

Now converting the combined matrix as

From this it is seen that whatever the values of the coordinates does not effect the value of the plane with equation as

So it is verified that the subspace of R3 such that it satisfies x+13y-3z=0 consists of all vectors.

going on the gulag might you to find a better answer for that problem

answer; /// i believe its (bas design);