Let A 3 3 0 6 1 1 3 3 3 3 9 9 1 0 1 2 8 2 4 6 rightarrowrre

Let A = [3 3 0 6 -1 1 3 3 -3 3 9 9 -1 0 -1 -2 8 2 -4 6] rightarrow^rref [1 0 0 0 0 1 0 0 0 3 0 0 0 0 1 0 4/3 -2 -2 0]; answer the following: a. Is b = [1 4 2 7] in ColA? Why or why not? b. Find a basis for ColA. c. Let beta be the basis for ColA from part (b). Find [x]_beta, the coordinate vector of x with respect to beta where x = [6 7 4 17] d. Find a basis for NulA. e. Based on part (d), give a non-zero vector in NulA. Let alpha be the basis for NulA and let [2 -3] be [x]_alpha, the coordinate vector of x with respect to alpha. Find x.

Solution

Solution : ( b )

Add (-1 * row1) to row2

Add (-2 * row1) to row4

Add (-3/2 * row2) to row3

Add (-5/2 * row2) to row4

Add (-1 * row3) 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 3 pivots (hilighted above in BOLD).
Because we have found pivots in columns 0, 1 and 3. 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 :

-------------------------------------------------------------------------------------------------------------------------------------------------------------

Solution : ( a ) No. By reduced Row Echelon Form, we get ;

---------------------------------------------------------------------------------------------------------------------------------------------------------

Solution : ( d )

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

Divide row1 by 3

Add (-3 * row1) to row2

Add (-6 * row1) to row4

Divide row2 by 2

Add (-3 * row2) to row3

Add (-5 * row2) to row4

Divide row3 by -5/2

Add (5/2 * row3) to row4

Add (-1/2 * row3) to row2

Add (1/3 * row3) to row1

Add (1/3 * row2) to row1

The matrix has 3 pivot columns (hilighted in BOLD) and 2 free columns; because the matrix has 3 pivots, the rank of the matrix is 3.

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 :

3	-1	-3	-1	8
0	2	6	1	-6
0	3	9	-1	-4
6	3	9	-2	6