Question

12. Find bases for Nul A and Col A. (8pts) WA 5 -3 1 1 1 - 1 2 22 5 0 - 8 - 2 4 -4 8 3 - 2

Answer 1

Given, A = -3 2 22 5 5 -2 -4 -8 8 -2 4 ook, AX -- 0 X in IR Nulla) n 84 96 log ( where, x= oood and o: suu system AX=0 in Augmented matrix of the -3 -1 5 5 2 0 22 -8 4 ५ 8 3 -2 Let up apply grow operation to convert grow reduced e chelon form. -3 -1/3 R 5 5 2 ador 22 -8 wis of ا -2 -2 oo 1 -93 - Yay R2-2R, 6870 173 1330 - 8 4 Rut 4 RI -2 5 20 3 -2% 1 - % R2 I 68 13 78 -2 -8 20/3 پا -23 1 -Y o Roz+ 8 R2 - Yu 1 o Ry - 20 R2 13768 94/17 -39717 0 17 Rz -Y 94 -73 73 Уч 13/64 -33/17 0

- Yng a - yn 74 0 - 0 RY- RIN Ru + 3 3 R -) 2 - 3 Уч Rita Rao s RittaRo -1/4 "/4 M(say) O 0 0 O This grow echelon form. This system is equivalence to 14 Ug = 0 Hy + 4 z = 0 2 = -4/4243 24. = 24 14 43 Hu zo so, Nul(a) = {"(*) : ax=0] o } Уч Hoz -74 ise, ER ER = span % -94 . Hence, Basis of Nul(A) is У, Уч 22- Col (A) = span { 2. O -4 8 8 ),(), ( , From Row echelon form of A i M= - 74 mu Basis {(1)(61). So, of Col(A) is