Question

(4) (15 marks) Repeat the Question 2 for the following matrices -3 4 0] 0 0 A -2 30 B 0 -1 0 -8 8 1 0 0 1 ū= 10 = > 3

Answer 1

0 A b= -2 0 li B = 0-1 The characteristic equation of A is 1 A-10 1-0 키 -3-1 4. O do-1,111 (Highs=0 -2 3rd 8 2 4 dal, (A+1)0 000 -2 4 0 1-8 8 2 -22+4y=0 razy & -4x+4y +3=03+42-44 2 → sy-ay= 4y=Z :, X, z = for Z=1 d=1, CA-I)x=0 -::00 ss = & 3=kto. X= *** [*] ". Let us take X2= 1:] Xz= [:] Detine diagonalize matrin P= poli 1pI+0. -X1 X2 X3 0 pa 2 4 4 they by Diagonalization p'ap=D POB as p'ap = -D

(VI) b= [] a [3]*1 [i]+c[] 2atb atb +4+с then _26+6=1 [band szatb=1 a+b=0>92-6 How → a=1 and 4a+C=3 Carl :. b = to) +(1) = 1 X, +(-1) X 2+ (DX3 * 4 A X₂ - Alo À 6= A[xx-x2 - x3] = * Alba (jox, - colox₂ - (1) xz = x,-Xz_x3 = Ab= [3] 2 ob HA (3) VIL by diagonalization P'Ap=1 = A= PDP" BAR A Aa CPD pul XPDPU) = (PD) (Ap) opta po pl similary pokpl ki is non negative integer AAE PDK pt 1 we have ikol KEN TO Vill Alo p b py 0:16:16:] [ ] - Gi]-+ Alor PIPY I A b= 1b = which is same as (VI). Hinge In This case pas PAP=B TX yes A is similar to B. P= X2 X, X3 faken