Question

3. [0.71/1.42 Points] DETAILS PREVIOUS ANSWERS POOLELINALG4 4.4.011.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Determine whether A is diagonalizable. 2 0 2 A = 0 2 2 2 2 0 Yes No Find an invertible matrix P and a diagonal matrix D such that p- AP = D. (Enter each matrix in the form [[row 1], [row 2], ...], where each row is a comma-separated list. If A is not diagonalizable, enter NO SOLUTION.) (D,P) = ( [(-2,0,0],[0,2,0],[0,0,4]][[1, - 1, – 1],[1, – 1,1],[1,2,0]] ) Need Help? Read It Master It Talk to a Tutor Submit Answer

Answer 1

Ano 2 2 2 2 2 2 0 first we find characterstic polynomial of A, CA (A) = 1 AI-AI- d-2 0 -2 O a-2 -2 - 2 - 2 a - l(a) = (-2) [ ald-2) –4] -0 -2 [0 +2 (2-2)] = (2-2) (2²-22-4) - 4(2-2) =a²_22²-22²+ua-u2+8 -4A+8 = 2²-ud²-4atto = ?(2-4) -4 (2-4) (a²-4) (2-4) (Ala) (2-2) (+2) (2-0) Then eigenvalue d = = 2,-2,4 since all eigen values are distinct matrix A i diagonalizable. For an Invertible matrix P and a diagonal matsix D P-AP =D we know that I has diagonal entries as eigen values and phas entries eigen vectors as column vectoas. -2 D o 4 -1 so, +2 0

A n = ax 2 2 2 13---19) 2 2 20+20 u 2b +26 -2a -2b - 24 20+2b 20+21=-2a 26t2=-26 20+2b =-26 Ya+2c=0 ]= 49-4620 46 +2c=0 Ta=b 2at 2b +2l=0 ua +21=0 9 TC= - 2a [2a=26=-c (= -2 Then eigen vector consponding to d=-2 is let a=1 > b=t, [1] - 2 (il d = 2 = li i fjli)--[] 2 atzi 2b +2. 2a + 2b - L 24 2b 20 20+21=2a 26+21=26 2a + 2b =26 [c=0 & Ic=0 a=-b let a=t then b=1 eigenvector corte ooresponding to da 2 is Then ]

(iii) d=43 o : : :11}]-4 (8) 9 20+20 2 b +20 za teb ub чC 20+2c=4a 2b +26=ub zat 2b = 40 c=9 g Ta=b=c 다. cab Then, let azt a bat, cat sponding to d=4 is L eigen vetor cossos 1 Then, st I 1 (1 1 -1 1 -2 1