Question

Question 3. (20 pts) Let A= -3 9-27 2 -6 4 8 3 -9 -2 2 Find a basis for Col(A) and a basis for Nul(A).

Answer 1

③ The given matrice is A= -3 % 2 - 6 -2 1 + 8 2 - -2 2 we appely elemen fany To O operation 0 1 . RI 2. R2 PR 6 g 4 & - 2 1 2 3 - 2 2L Riz Ž R 3 g 2 41 - 2 1 -2 2 -3 3 - R2² 3 Ry TR2 -3 2 1 4 19 คน 2 3R-R3 0 0 8 | 6 R3² R3-2R22 0 0 0 2 1 4 19 0 - 29 1 C2 - C, -24 (23) 0 1 14 0 18 O -2% () 0

chalaga O 4 19 0-28 O C3=43 R2 = 28 R2 0 119 0 o 0 e4 = 4 44-194 $ 0 EB, 0 0 B is a matrix. no w echelon det X=(x, yz, w) € Nel (A) then Axao 2) BX2O. N 0 20. w 2) nty Z W 2 2) aty.20 Zzo W 2 O. 2 ) 22-7 Z 20,W20, L, X (?) (99-03)

Therefore Nuu(A) = 1 se($):CER Since Row echelon matrine B Contains beading 1 in 1st, 3rd 4th colunan then and 7 colcas a spam? (3), (3, (0) ao 3 i e basis for muu (A) is and basis for colcA) is 3 2 2 4

41) rank (A) = number of element in basis for Col(A) 3 dim (Nul(A)) a number of element in basis for NUICA) I. dim (.col (A)) = number of element in basis for colcA) 23. 37 Number of Column=4. dim (Mul (A)) + dim ( 201 (A)). 1 +3 24 - dim (MUU (A)) + dim ( COICA)) = Number of Column of A.