Question

PLZ SOLVE BOTH WRONG PARTS

For the matrix A, find (if possible) a nonsingular matrix P such that p-1AP is diagonal. (If not possible, enter IMPOSSIBLE.) A = 1 -2 P = 4 1 11 Verify that p-1AP is a diagonal matrix with the eigenvalues on the main diagonal. 1 0 p-1AP = 0 3 X Need Help? Read It Watch It Talk to a Tutor [1/2 Points] DETAILS PREVIOUS ANSWERS LARLINALG8 7.2.009. For the matrix A, find (if possible) a nonsingular matrix P such that p-1AP is diagonal. (If not possible, enter IMPOSSIBLE.) 2 -2 3 A = 0 3 -2 0-1 2 -1 1 7/2 P= 1 0 -2 1 0 1 1 Verify that p-1AP is a diagonal matrix with the eigenvalues on the main diagonal. 2 -2 3 0 3 -2 p-1 AP = 0 -1 2 11

Answer 1

# A= [?] The characteristic. polynomial are 2,1- 92-92 +8 -8 = 0 → 4 (1-3) = 0 8-7 - 2 - 4 1-7 X X a = 0,9

The eigen natues 1,=0, A2=9 a=0 8 [:] [3] [:)] 82 – 29 ,-4x+y = 0 If we take a=1, then y=4 The rector 1 4 eigen rector 9 - 9 (6) E :] [:] [:] -2 -2 y = 0 -42-8y = 0 If we take y = 1 then 2=-2 The eigen vector 1 2 (1) The eigen values a=0 the corres vector responding eigen (6) The eigen values ?2-9 the con e Contestanding eigen recton (-3). 1 - - 2 Pe 4 1

IPIS 1 +8=9 Adj P: 14-21 fa ) 5 (41) '. Adi P PL 219 19 19 -4/9 yo 2/9 -4/9 29 P'AP - - -690)6 ) - (83) 11 9 Cans) 2 # A= - 2 فر) 3 -2 0 -1 2 The characteristic polynomial are - 2 3 2-7 0 0 3-7-2 -1 2-3 -> (2-2) (a2+2 +6 -2) + 2 (0) +3 (0) - 0 -> C2-9) CA2_52 + 4) = 0 >> (2-a) (2²_42 9 +4)=0 - (2-2)[^ (1-4) -1(9-4)-0

> (2-2) (9-4) (9-1)=0 :.2= 1, 2,4 The eigen values 2,= 1,92-2, 23 = 4 2=1 -2 3 X ܘܘ ܘ o Y 0 2 -2 0 - 1 1, .. 2-29+37= 0, 24-22=0, -y +7=0 If we take z=1 theny X=-1 The vectoras 1 1 a = 2 &] - 2 3 z o 1 - 2 y 0 0 0 Z -1 0 2y + 32=0, 7-22=0, - Y=0 If we take a then y=0, Z=0 The rector 1 o o a = 4 ) 2 -2 3 1 O 1-2 -1 -2 y 2 0 0 -y-22=0 -2x - 2y + 3a = 0, If we take q=1 , then y = -2,2=772

The eigen vector 7/2 -2 The eigen value a =1 then corresponding eigen vector 1 The eigen value a=2 then corresponding eigen vectos 1 0 The eigen nalue 9 = 4 thens Corores 4 then corous bonding eigen 7/2 nector 2 1 1 7/2 1 0 2 1 0 1 1 7/2 1P13 C 1 0 -2 0 1 = -1x0 -1 (1+2)+7/2(0) -3 | 7/2 2 adjp- (+)?! -15%/ +15 -11-21 +174-1773) +/18 | - | | + 1 1 - 1 1 1 0

- o T- I P = adj 0 E 3 lo-1373/ -1 - 2 - 3 -% -3/ -1 -2 -3 -972 3/2 1PL 1 -1 2 3/2 - 1/2 o - 1/3 13 o 1/3 2/3 1-1 1 7/2 3/ - / 0 3-2 1 0 -2 1 0 1 013 13 2/3 1 1 7/2 2 2 -1 0 -43 473 1 1 2-2 3 z PAPE 0 -12 =( - = 0 - 2 T 0 = -1 2 2 bo (Ans)