lin Consider the 3 x 4 matrixA 1 2 4 5 0 7 3 4 1 7 6 2 Find

Consider the 3 x 4 matrix.A = (1 2 4 5 0 7 3 -4 -1 7 -6 2) Find Col A. Determine whether the vector v = (1 - 2 3) belongs to ColA Find NullA. Determine whether the vector u = (1 2 - 1 3) belongs to Find a basis for NullA. Find a basis for Col A. Find rank A and nullity A.

Solution

COLUMN SPACE

Add (-2 * row1) to row2

Add (-4 * row1) to row3

Add (-13/10 * row2) to row3

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 yellow)
Because we have found pivots in columns 0 and 1. 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:

ROW SPACE

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

Add (-2 * row1) to row2

Add (-4 * row1) to row3

Divide row2 by -10

Add (13 * row2) to row3

Add (-5 * 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 yellow comprise a Basis for the Row Space of our matrix:

NULL SPACE

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

Add (-2 * row1) to row2

Add (-4 * row1) to row3

Divide row2 by -10

Add (13 * row2) to row3

Add (-5 * row2) to row1

The matrix has 2 pivot columns (hilighted in yellow) 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 yellow)...

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:

The rank of the matrix is 2. It equals the number of leading entries.

The nullity of the matrix is 2. This is the dimension of the null space. It equals the number of columns without leading entries.

1	5	3	7
0	-10	-10	-20
4	7	-1	2