Question

Let To 1 3 3 0 -1 3 2 2

(a) Use Mathematica to find an echelon form of A.

(b) Using your answer to the previous part, find the rank and nullity of A.

(c) Find a basis for the row space of A.

(d) Find a basis for the column space of A.

(e) Find a basis for the null space of A.

Answer 1

(a) solution: reduced row space echelon form of A

0 1 -1 3 3 0 3 2 11 0 -2 41-1 0-1 3 2 2 Make zeros in collumn 2 eacept entry at ow 1.column 2ovot enby 0 1 3 3 0 2 41 1 1 3 3 01 22 0 02 55 0 -1 3 2 2 0 1 -1 S 3 0012 2 002 0 02 55 0105 5 0 01 22 0 02 5 0 025 5 0 0122 Subtract now 2 mupled by 2 trom sow (R-R-(2) R) 0105 5 0 01 2 2 Suttract row 3 mutplod by 3 from oW(R -RL (R) 010 0 0 0 01 22 0 001 1 utsract ron 3mutplnd by 2 from ron 2R R (2)R 0 1 0 0 0 0 01 L 0 10 0 0 0100 0 0 01 0 0 Anowen troA)-

(b) rank

Solution: Answer. Since there is 3 non-zero rows, then Rank(A) 3

nullity:

rankA + nullityA = number of columns of A.

therefore, nullityA= no. of columnsA - rankA = 5-3 =2

(c) row space of A

First, we must convert the matrix to reduced row echelon form: Add (-3 row1) to row2 0 1-13 3 00 12 2 0-13 2 2 Add (2 row1) to row3 0 1-13 3 001 2 2 00255 0-13 2 2 Add (1 row1) to row4 0113 3 00122 002 5 5 Add (-2 row2) to rows 0 1-133 00 12 2 00011 00255 Add (2 row2) to row4 01-133 00 12 2 0001 1 0001 1 Add (-1 row3) to row4 01-133 001 2 2 0001 1 000l0 0 Add (-2 row3) to row2 01-133 00 10 0 0 00 11 0 Add (-3 row3) to rowl 01-100 00 100 00011 00bl00 Add (1 * 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

(d) column space of A

Add (-3 * row1) to row2 01-1 33 00122 0-24-1-1 Add (2 * row1) to row3 01-1 3 3 0012 2 002 5 5 0-13 2 2 Add (1 * row1) to row4 001 2 2 002 5 5 00 2 5 5 Add (-2 row2) to row3 01-13 3 001 2 2 000 1 1 Add (-2 * row2) to row4 01-1 3 3 00 12 2 000 11 000 11 Add (-1 row3) to row4 01-13 3 0012 2 00011 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); 0 1-13 3 0101122 The matrix has 3 pivots (hilighted above in yellow) Because we have found pivots in columns 1, 2 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: A, B and C are any real numbers. 4

(e) null space of A

First, let's put our matrix in Reduced Row Eschelon Form.. The matrix has 3 pivot columns (hilighted in yellow) 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 yellow) and turn it into its own matrix: Let's multiply this matrix by -1: 0-1 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): 0-1 Finally, the Null Space of our matrix is defined by scalar multiples of these column vectors A 10+B 10 -1 A and B are any real numbers.