To determine whether the set S spans R3, we need to check if the vectors in S are linearly independent and if they can form any vector in R3.

The set S = {(1, 0, 3), (2, 0, -1), (4, 0, 5), (2, 0, 6)} contains four vectors in R3.

First, we can see that all vectors have the second component as 0. This means that they all lie in the plane x-y=0 in R3. So, they cannot span R3 because they cannot form a vector that has a non-zero y-component.

Second, we can check if these vectors are linearly independent. We can form a matrix with these vectors and then reduce it to row-echelon form.

The matrix is:

1 0 3

2 0 -1

4 0 5

2 0 6

We can subtract the first row from the second and the fourth rows, and subtract twice the first row from the third row to get:

1 0 3

1 0 -4

2 0 -1

1 0 3

We can see that the first and the fourth rows are identical, and the second row is just the negative of the third row. This means that the vectors are linearly dependent.

So, the set S does not span R3. The subspace that it does span is the plane x-y=0 in R3.

Question

To determine whether the set S spans R3, we need to check if the vectors in S are linearly independent and if they can form any vector in R3.

The set S = {(1, 0, 3), (2, 0, -1), (4, 0, 5), (2, 0, 6)} contains four vectors in R3.

First, we can see that all vectors have the second component as 0. This means that they all lie in the plane x-y=0 in R3. So, they cannot span R3 because they cannot form a vector that has a non-zero y-component.

Second, we can check if these vectors are linearly independent. We can form a matrix with these vectors and then reduce it to row-echelon form.

The matrix is:

1 0 3

2 0 -1

4 0 5

2 0 6

We can subtract the first row from the second and the fourth rows, and subtract twice the first row from the third row to get:

1 0 3

1 0 -4

2 0 -1

1 0 3

We can see that the first and the fourth rows are identical, and the second row is just the negative of the third row. This means that the vectors are linearly dependent.

So, the set S does not span R3. The subspace that it does span is the plane x-y=0 in R3.

Knowee AI · Accepted Answer

To determine whether the set S spans R3, we need to check if the vectors in S are linearly independent and if they can form any vector in R3.

The set S = {(1, 0, 3), (2, 0, -1), (4, 0, 5), (2, 0, 6)} contains four vectors in R3.

First, we can see that all vectors have the second component as 0. This means that they all lie in the plane x-y=0 in R3. So, they cannot span R3 because they cannot form a vector that has a non-zero y-component.

Second, we can check if these vectors are linearly independent. We can form a matrix with these vectors and then reduce it to row-echelon form.

The matrix is:

1 0 3

2 0 -1

4 0 5

2 0 6

We can subtract the first row from the second and the fourth rows, and subtract twice the first row from the third row to get:

1 0 3

1 0 -4

2 0 -1

1 0 3

We can see that the first and the fourth rows are identical, and the second row is just the negative of the third row. This means that the vectors are linearly dependent.

So, the set S does not span R3. The subspace that it does span is the plane x-y=0 in R3.

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, 0, 3), (2, 0, −1), (4, 0, 5), (2, 0, 6)}

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