Question

linear algebra 2 part mcq

r(A) Find and n(A) A = 1 - 3 4 -1 9 -2 6 -6 -1 -10 -39 -6 -6 -3 3 -94 9 0 a. r(A) = 5 n(A) = 0 b. r(A) = 3 n(A) = 2 c. (A) = 0 n(A) = 5 d. r(A) = 1 n(A) = 4 e. r(A) = 4 n(A) = 1 f. r(A) = 2 n(A) = 3

part a

part b

Answer 1

1 -3 4 4 - 9 A= 2 -6 o 2 -3 - 1 97 R2 + R2 + 2 R1 -10 R3+R3 +3R, -6 Rat Ra-BR1 9 0 -3 9 o 6 ao a -9 24 w W -9 4 o - 8 12 -27 1 9 1 Rot Rs-2R2 5 -7 R3 R₃ - 3R2 Rat R4 +4R2 ooo w + NOO 8 0 2 ņoo R3 R4 ooo w O моо 5 1 -3 0 5 O ☺ -3.5 0 (RREF) o o o o оо o there 3 non eso now I &CAJ=3 there are 2 non pirat collemn & nlas=2 b. &CA)=3 n(A)=2 -> Ang (2) A= 1 3. So 4 eigenvalue is IA-as1=0 2 l-e. 1-9 14 3 12-30-10 = (1-5) (2+2) =0 4 d=5, -2 l-e. eigenvalues are 5 and 2 eigenvector for de 5 l-e. [A-51 [ 73] ~0-34] Λοω sebre [ ]UJ-0 is free let ug=t then ve=314t so 3/4 314 ( 3 is eigenvector for da 5 if t = 4

Now eigenvector for 2=-2 [A-(-2)I] = Ug is free lel uzat than ug = - t so [Jt; 0) is eigenvector for 2=-2 ift=1 sol D = 5 3 and p = 0 -2 4 ܢܐ d. A= PDP-1 3 P = 13 182) Ang 4