Question

Find a basis for the subspace of R3 spanned by S. S = {(1, 4, 6), (-1, 5, 6), (2, 5, 1)} lll -001 11 Show that the set is linearly dependent by finding a nontrivial linear combination of vectors in the set whose sum is the zero vector. (Use S1, S2, and s3, respectively, for the vectors in the set.) S = {(1, 2, 3, 4), (1, 0, 1, 2), (3, 8, 11, 14)} (0, 0, 0, 0) = Express the vector sz in the set as a linear combination of the vectors S, and sz. S3 4s, - S2 Determine whether the set S spans R3. If the set does not span R3, then give a geometric description of the subspace that it does span. S = {(1, -7,0), (0, 0, 1), (-1,7, 0)} S spans R3. O S does not span Rp. S spans a plane in R3. O s does not span RP. S spans a line in R3. O s does not span Rp. S spans a point in R3.

Answer 1

To find a basis for the subspace of R3 spanned by S, we need to determine if the vectors in S are linearly independent. Using matrix notation, we can set up a system of equations and solve for the unknown scalars. If the only solution is the trivial solution, then the vectors are linearly independent and form a basis.

Step-by-step explanation:

To find a basis for the subspace of ℝ3 spanned by S, we need to determine if the vectors in S are linearly independent. We can do this by setting up a system of equations where each vector in S is a linear combination of the unknown scalars. If the only solution to the system is the trivial solution (where all scalars are zero), then the vectors are linearly independent and form a basis.

Using matrix notation, we can write the system as:

[1 -1 2]x = 0

[4 5 5]y = 0

[6 6 1]z = 0

Solving this system of equations, we find that the only solution is x = y = z = 0. Therefore, the vectors in S are linearly independent and form a basis for the subspace of ℝ3 spanned by S.