Question

Find N(A) A= 1 -3 4 - 1 9 -2 6 -6 -1 -10 -39 -6 -6 -3 3 -94 9 0 oa. 1 -2 -3 3 - 3 6 9 N(A) = 4 -6 -1 -6 -1 -10 -6 9 9 0 3 0 N(A) 1 0 2 0 N(A) = 2 1 0 -2 0 d. 3 -10 0 N(A) 2 0 0 N(A) -2 6 -3 4 N(A) = -1 -6 -1 -10 9

Answer 1

Given. A- 4 -19 6 -6 - to -6 -6-3 3 g 4 9 0 -3 Now we find row reduced echelon form of A. 1 -3 А R2+2R, R3+3R1 R4-3RA -3 8 -924 6 o ao оо 12-271 5 -7 RĄ -QR2 R3-3R2 R444R2 Oo oow o on 0 o 5 R34 0 qe 0 roo 0 0 م 0 00 تراحمه O_ܢܺܝ 1/2 R2 / R3 o w 0 O 0 Now we Know that NIA) MIRROPA)) i.e. M(A) = N(B), nu ENCA) na NS Then BX=0

OS дуу ч о о о 15. Nе, и le. - 3 *, This gives, х, - 3к, + SX4 + 1 xs 20 ч А. - 4*- со ng Now, put put mas ao above a equation, L, Уи,_+ SX4 = 0 2х3 , х. -4 E) : ", ". ЗА امر с as = 0 Now, 3х5х4 уу. -. Х. 4 % 름 КА 4 ms -5 k, 39 1 D x2

so we get, 3 1(B) d 0 onwo O since N(B) = N(A). There for the correct option is