Question

(21) (15 marks). Given 1 A 1 3 1 3 4 0 0 1 1 0 0 2 2 2 0 0 3 3 3 (a) (5 marks). Find a basis for N(A) (null space of A). (b) (5 marks). Find the rank of A; (c) (5 marks). Find a basis for the column space of A.

Answer 1

Given 13 4 - 1 4) Ma 0 0 2 2 2 0 0 3 3 3 3 0 2 3 ܂ Rit Rr-R₂ R₃-) Rz=R2 0 0 R4>RA-BR 0 0 0 0 0 FOX Null space of A Ax=0 2 3 - 1 0 X2 0 0 1 X3 44 0 0 0 0 0 0 0 0 0 0 Xr ܢ- Free variables X21X4 4XG det X2=a ,X4=b, Xs=0 X,+32 +26+3 C = 0 X3+b+C = 0 X = -39-2b-3c Xz=-b-c NCA) is given by [-Ba-2b-36 a 10-1 -b-c 1 b C

,'. basis of NCA) - -3 -2 0 0 TO c510910 ) Llo Dimension of NCA) is = 3 pour b) can write Pivots are in the first and third COLUMNS of the echelon form, we original matrise Of column id3 as column space bai .. basis for column space of NCA) 70 is Rank of A is 2.