Question

Question 3. (20 pts) Let A= -3 9-27 2 -6 48 3 -9 -2 2 Question 4. (15 pts) Let the matrix A be the same as in Question 3 (1). Find the rank of A (2). Find the dimensions of Nul(A) and (0) (3). How do the dimensions of Nul(A) and Call relate to the mber of columns of A?

Answer 1

Questions 3 7 Let A = 2 9 -6 -g -2 4 -2 8 3 2 now he apply elementary row operation on A. 그 -3 2 3 2 -6 R2 2 -6 -2 7 48 2 4 8 -2 7 -2 2 -3 g 3 - ŹR RzTR2 :3 g -2 7 -4 g 3 -9 -2 1 2 R2 +3R, 2 4 -3 Rz +R2 moo A 19 oo 1 -3 24 o 4 tg O 28 ㅗ OO - Rz 128 R3 94 too -4 ㅗ 3 2 1 R, 4R3 R2 - 19/ R1-2R2 D). ㅗ No of Don Zero row is 3 so rank of A=3. matria D. then C € Null CAD ☺ 2 How from redued ܩ b d Ax=0 hold Dx = 0 a-3 b=0 @=0 d=0 2 then real number take t be b=t a=3t. t 3t 25 so the solution Ooru therefore basis for Null(A) = { Oorw

Зо dimentions of Niel (A)=t. Column or again the reduced matrix D has hree pivot dimention three Column contain pivot element. So dimentions of col (A) = 3. therefore dim Nul(A)=1, and dim ColCA) = 3. and the Relation between dementions of Nul (A) and ColCA) of columns number is, af dimention of Nul(A) + dimention of COP (A)= number of column of A. dim Nul CA) + dim colCA) = no. no.es of column in A. or