Question

no calculator please

2. (12 pts) Given the matrix in a R R-E form: [1 0 0 0 3 -2 1 1 0 0 0 0 0 1 0 0 0 0 0 0 (a) (6 pts) Find rank(A) and nullity(A), and nullity(AT). (b) (2 pts) Find a basis for the row space of A. (c) (2 pts) Find a basis for the column space of A. (d) (2 pts) Find a basis for the null space of A.

Answer 1

2. 0 3 2 O 0 O O O 14x6 a> from the above RR-E form 3 rank (A)= nullity (A) number of column pank(A) 6 - 3 = 3 0 T A o 0 3 - 2 10 - 2 0 6x4 we know, rank (A)= rank k (AT) irank (AT) = 3 - nulity (AT) = number of column pank (AT) 9-3 = z 1

(6) Row space of A is generated by (1,0,0,0, 3,-2), (0,1,1,0, -2,1), (0,0,0,1,1,1) and (0,0,0, consider 4 (1,0,0,0,3, -2) + C2 (0,1,1,0,.2,1) tcz (0,0,0,1,1,1) = (0,0,0,0,0,0) Now, and Hence the vectors (1,0,0,0,3,-2) (0,1,1,0,-2, 1) (0,0,0; 1, 2, 1) are Linearly independent. these forms a basis for So pow space of A. the } basis set for the row space of A is a (1,0,0,0,3, -2), (0, 1, 120,-21) (0,0,0,1,1,1) columns space of a is generated by the columns of A. generated by the vectors - (1; 0,0,0), (0,1,0,0), (0,1,0,0),(0,0, 1,0) Ĉ3,-22 1,0), (~2,181,0). That

{(1,0,0,0), (1,0,0,0), (0,1,0,0) basis for for the column space the set forms of A (0, 0, 1,0)) (d) for finding null space of A a basis for the Consider 4 12 AX=0, where x = 13 nu us AX=0 2.6 o 0 3 M o -2 av uz nu o 0 O O 0 0 "L 26 0 >>> 4 +305 - 206 20 12+13 - 2u5 +46 nutus tu6=0 Take W6 ki Us=kz Then nu and - k-k2 in = -3K2+2K K3 H2 = -K3+2 K₂ -ki Tave uz then

. (hrl2 ,U3, U4, KS, ko = (-3K2+2 kg -K3+2 K2 - kg K3g - 4-k2 g K2 ,k) V (2 Z ll K, (2,-1,0,1,0,1) + Kg (-3,2,0, 1, 1,0) +63(0,1,1,0,0,0) arbitnery ki K2, Kz are arbit Constant. also, K,K2 o K3 ER by and Nel space of A iš generated (2,1,0,1,0,1),(-3,2,0,-1,1,0) (0,-1,1,0,0,0) since, dim of nul space of A is 3: Nullity of a 23 This set forms a basis for Null space basis for mell space of a (2,-1,0,1,0,1),(-3, 2,0,1,10), is a (0,1,1,0,0,0) That i 2