Question

Use the fact that matrices A and B are row-equivalent. 1 2 1 0 0 2 5 1 1 0 3 7 2 2 -2 5 11 4-1 4 1 0 30-4 0 1 -1 0 BE 2 0 0 0 1 -2 0 0 0 0 0 (a) Find the rank and nullity of A. rank nullity (b) Find a basis for the nullspace of A. (c) Find a basis for the row space of A. III 100- DUL III 11 (d) Find a basis for the column space of A.

Answer 1

Q given that A & B dow equivalent - 2 - O - 2 5 Az 3 7 يع 2 -2 5 5 4 -) / -4 0 3 - -) O 2 -- 0 B = - 2 - 0 0 o o (a) لم) rank (A) - rank (B) By rank mullity theorem rank (A) + nullity (A) = no. of Column in A 5 + mullity (A) = 介个 nullity (A) = 2 (6) let x E Null (A) T xo] xu SO 2 = x, xn x₂ then AX => Ba -

3 о 6- S) xa — п -) о 1. п ราม о nx о о де So 2, 4 ox. + 32, + oxy го *zh — О. п. 3 + oxy + x б ( hx - 2 х let у с x3 = = { д then хи 11 7 — 8 – 24 2 x x₂ £х — — — 38 + ч — 33 sont (7 - 38 int 2 , 7. п 8 Ч 11 ux sx — 1

-3 u x, -2 ra tt x = X3 0 2 ry 25 then Baris of mullspace (Al is 3 -2 0 B - Null (A) دہ 0 o now (c) of B T rows mon & TO I, zero => they construct the besis of row space of Row (A) is 3 O -4) o В. -) 2 Row (A) [o [ -2] O ro O

(d) I, I I & IV care pivot column in B => I, I & IV Column of matrix colum space of are contructed the Basis of Col (A) is 2. 5 no B Col(A) 7 3 -) ) 5