The reduced row echelon form of A 3 7 5 4 9 21 15 12 1 4 2

The reduced row echelon form of A = [-3 7 5 4 9 -21 -15 -12 1 -4 2 -3 -11 19 33 8] is R = [1 0 0 0 -3 0 0 0 0 1 0 0 5 4 0 0] Consider W = Span(S), where S = {, , , )} Find a basis for W, consisting of vectors from S. What is the dimension of W? Express the other vectors from S as linear combinations of vectors from the basis. Find the rowspace, columnspace and nullspace of A.

Solution

Solution : 8)

C )

Row Space of A :

First, we must convert the matrix to reduced row echelon form:

Divide row1 by -3

Add (-7 * row1) to row2

Add (-5 * row1) to row3

Add (-4 * row1) to row4

Divide row2 by -5/3

Add (-11/3 * row2) to row3

Add (5/3 * row2) to row4

Add (1/3 * row2) to row1

Because we have only performed linear operations on rows, the non-zero rows in the reduced row echelon form of the matrix comprise a Basis for the Row Space of the matrix.
(Note that this is not true of the Column Space; the Column Space certainly changes as you perform row operations.)
The rows highlighted below in BOLD comprise a Basis for the Row Space of our matrix :

and now ;

Column Space of A :

Add (7/3 * row1) to row2

Add (5/3 * row1) to row3

Add (4/3 * row1) to row4

Add (11/5 * row2) to row3

Add (-1 * row2) to row4

First, we must reduce the matrix so we can calculate the pivots of the matrix (note that we are reducing to row echelon form, not reduced row echelon form):

The matrix has 2 pivots (hilighted above in BOLD)
Because we have found pivots in columns 0 and 2. We know that these columns in the original matrix define the Column Space of the matrix.
Therefore, the Column Space is given by the following equation:

and now ;

Null Space of A :

First, let\'s put our matrix in Reduced Row Eschelon Form...

Divide row1 by -3

Add (-7 * row1) to row2

Add (-5 * row1) to row3

Add (-4 * row1) to row4

Divide row2 by -5/3

Add (-11/3 * row2) to row3

Add (5/3 * row2) to row4

Add (1/3 * row2) to row1

The matrix has 2 pivot columns (hilighted in BOLD) and 2 free columns; because the matrix has 2 pivots, the rank of the matrix is 2.

Let\'s take the \'free\' part of the reduced row echelon form matrix (hilighted below in BOLD)...

and turn it into its own matrix:

Let\'s multiply this matrix by -1:

Now, we add the Identity Matrix to the rows in our new matrix which correspond to the \'free\' columns in the original matrix, making sure our number of rows equals the number of columns in the original matrix (otherwise, we couldn\'t multiply the original matrix against our new matrix):

Finally, the Null Space of our matrix is defined by scalar multiples of these column vectors:

1	-3	-1/3	11/3
7	-21	-4	19
5	-15	2	33
4	-12	-3	8