Question

8. [0/5 Points] DETAILS PREVIOUS ANSWERS LARLINALG8 7.1.021. Find the characteristic equation and the eigenvalues (and corresponding eigenvectors) of the matrix. 2 -2 0 3 -2 0 - 1 2 (a) the characteristic equation (2 – 2)(a – 4)(a − 1) X (b) the eigenvalues (Enter your answers from smallest to largest.) (11,12,13) = ( (1,2,4) the corresponding eigenvectors x1 = (1, - 2,9) X2 = (0,2, - 2) x X3 = (0, – 1,1) x
please solve both as other did wrong plz

Answer 1

8 matix -2. a O 3 -2 0 2 a) The characteristic equation is JA-AI)20 2- 9 0 3-1 -2 -0 0 2-2 ne get Expand along wlumn I (2-d) { (3-2)(2-0) -2] =0 (2-x)(x² - 5x+4)=0 (x-2)(x-1(d-4)=0 (-1(A-2) (-4)=0 b) The eigenvalue are (d, dz, dz) = (1, 2, 4) 9 sind une eigen westou for x=1 12-1 9 -2 1 -2 9 0 3-1 -2 0 0 2 -2 0 0 2-1 D - -1 0 soire fone malix equation we get 00 1 0 o 0 0 weget 24 +113=0 - 13 = 0 av = -13 + y = ₂

1 0 - . = A - X = (-1, 1, D CM 4= 2. 2-2 b 0 - 2 O 0 3-2 N 0 1 0 0 0 0 2-2 To -) 0 다. 0 0 M 월드, 1320 - 192 0 11) 122 0 A 0 이 (L, 0) 0) for a 니 29 -2 b 2 -2 9 13/ 0①0 0 이 3~4 0 0 - 2 2 2-4 0 이 - 2 0 0 0 eg 부를 병들이기 일드를 1년 -이 V. e-209 월 428 132 11 0 는 2. -22 원 - Li 도 라

we have matix 블 A- 4 -1 ៗ a) sind ure eigenvalues of A. TA-AI)20 4-1 1 7-1 2 (4-22 (7-2) +2=0 12- ||d+30 20 (2.5)(1-6)= 0 The eigenvalues are (d, d) =(5,6) b) sind the eigenreciou and then eigenspaces: fou =5 54-5 0 we get V +12=0 3 v = -₂ -1 F V2 B={1}} a B = [ { (-1,113 for dyle 6 4-6 11 2 1-6 -2 2 1 0 D we get , t 1 = 0 ,- 1 s

-½ -/ 12 Bz 3 Jos - {"}} B-U11) O:4:1 c) Now, BY here, number of eigenrecton (say, 2) is equal to the number of columns (say, 2) is mawily A. So, mattix A u diagonalizable such that D = PAP where D = 5 0 -1/2 P= 0 6 F Thus, it follows that [5:] mattis apresentation for I worrit the basis B = {x, 2} [2,10-1 and L] - P is the transition maltix (change of basis) from s to B. where -2 [ and these Ala 6