Question

Question 3 [10 points] Consider the following matrix A and its reduced row-echelon form: A = [-3 3 6 12 0 151 | 1 -1 -2 -4 0 -5 -6 3 9 15 12 18 rret(A) |-1 -1 0 -2 8 -3 [1 0 -1 -1 -4 -1] 01 1 3 -4 4 0 0 0 0 0 0 0 0 0 0 0 0 Find the dimensions of row(A), null(A), and col(A), and give a basis for each of ther Dimension of row(A): 1 Basis for row(A): 0 Dimension of null(A): 1 Basis for null(A): {|0| Dimension of col(A): 1 Basis for col(A): 101

Answer 1

3 3 -1 3 6 -2 9 12 -4 15 0 0 12 15 -5 12 1-6 v echelon form i 0 - -1 -4 - 7 ③ Non-zero row of matrix -B axe the basis for tow A. Bari e for row (A) = { [ 0 1 -1 -4 -1], [0 11 3-4 47 ? dimension of (row (A)) = 2 Nill space lit p be am matrid. the null epau is the ect of solution to the homogeneous system. Pžao N(P) & FRER" : Arno - Bawe know AXI equavalant to BX=0

Го - от го - тоо т зоо І Жі - 25 -— 4-%s - со Ҳ + ҳ» гэх ц* * * * х , tx4 14xs txt қоя - ку - 2x4 rчх, -цъс Basic vastakle free variable ГХtX Tч Хs 4X -Х2+5х4 +4X - X; 1 тоо-о. so Basic por null (A) oo o

dimention of mull (A) = 4 definition of Column (A) space two cretizia of column space. matrix must apan the entire spante set of all at o linear combination of the bacalumn actoruts must be limarly independent 1-6 3 9 15 12 18 T Calund span 737 01- 11- sorr157 How now r! echelon form 0 - 4 -4 - © Now linearly independent calumn - Pivot:calinn basis for calum A. hem we have calum ter and pineb column. n.37 so Baric for columns fr dimention of colump Ac 2.