Question

Diagonalize    

	a.
	b.
	c.
	d.
	e.
	f.

Diagonalize A A = 1 3 4 2 a. A = PDP-1 b. A = PDP-1 1 Р 1 1 OC. A = PDP-1 -1 3 P = 2 5 d. A = PDP-1 -3 1 P= -4 1 e. A = PDP-1 1 -1 P 3 1 Of A = PDP-1 P-[31] -- [6-2] [37] - [64] P=[ +3 z] --[: = D = 10 03

Answer 1

(1) Ans Given to [4 2] for eigenvalues of A, let | A-/1120 1-) co 2x h 2-) =) (1) (2)-12 zo =) 2-x-2x+x²_1220 12-31-10 =0 T → 12-5X +2 -1020 =) 1(2-5)+2(8-5) 20 z) (1-5)(+2) co G î = 5,-2 E-2 be the eigenvector corresponding 2 Let X = 5 ^= 5 and Let X= [mm] the eigenvalus î= 5, then (A-1) XI 2-0

(2) (A-SI) / 0 -S 3 2-5 니 3 아, 다. ]] []-[] [8] 니 23 디 44 +312 현 44 -3N2 on compairing and God - 342 20 + 3 12.20 which gives 4개는 3만 tm 기고래 업 3 Hence 세 5 be Let X2= C [] [] the ergon vector Corresponding the eigenvalue 2=-2, then (A +21 ) X2ző 게 (+2 3 [][] 4 242

(3) 130316 4 4 u . 374 +372 421 +4012' On Compairing um: [•] 3n+ 3n2's and 400+ una co + which gives 1 uc of x'=1 then aizal Hence X2= [1] Then the invertible matix P Corresponding the eigenvectors Xi and X2 is 3 -1 P= [ - 4 Naw IPL= 3+4= 7 Now Þ!= 1, 5122-4, Þ21=-(-1)-1, 522= 3 Then biz -4 BE ]= [! 3] Lbz1 Þ22 3

(4) Νσυ adj P= transpose of B 1 ] - 3 New Fe 1 i_adi (P) IP Te h 3 Now J - (+3 라마 - 2 FAR L LILLS] 1] - II [Ats JITS Eat [ ] 7부 مردم -2 15 420 2-2 -60 460 0

(5) Pap= To :]=0 = ) Ť Ap=D =) Plē'ap) ple ppp =) (pp)APP) - POP -) IAI = PDPT A=PDP? where pa 3 -1 and D=15 o o [ [o AS -2 o Option (a) 15 Correet.