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.

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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

(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) t

(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

(e) null space of A

First, lets put our matrix in Reduced Row Eschelon Form.. The matrix has 3 pivot columns (hilighted in yellow) and 2 free co

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