Question

please do number 3 only, thank you

In Exercises 2-9 find a matrix P that orthogonally diagonalizes A, and determine p-AP 3. A = 213 ]

Answer 1

Sel": BS Guien matix is — A = (3%). The eigen values of A are given by - 1A-111=0 (3-1)3-1 = 0 » X-01 +8 =0 3) -21-41+8=0 = (x-2)(x-4)=0 =) , = 2, 12 = 4 so, eigen values of A are a and a. Let v n eigen vector corresponding to 1,d. Then, (A-21), =0 =) ( :90) = (8) - atb = 0 so, v = (-1). Agaeis, let V2 = () be an eigen vector corresponding to X2=4. Then, (A-4I) V2 = 0 » ('-1)(a) = (0)

- -+d=o > c=d. so, vz = ('). " Matrix P that diagonalizes A orthogonally is - P = (viva) 6 :). Novo, per poi (5) -=(1-33-6-3 i pulap = *)(*)63) = D , diagonal form of a with eigen vallces as hal elements. Thew, p'ap = 0, 0-15%). 3) The giren matix is -

The eigen values of A are given by - 1A-1I[=0 3 16- 2V3 -0 12V3 71 =) (6-4) (7-1) - 12 = 0 > -131+42-12 =0 =) 2-131+ 30 =0 -101-31+30 =0 = (1-3) (7-10)=0 =) di = 3, 12 = 10 , So, eigenvalues of A are 3 and lo. Now , valet Vi = ( 8) be an eigen rectos corresponding to l. =3. Then - (A-3 I)vi = 0 > () (9) - lova 1463) () =(8) [R, 9 Ryz] lo 2013 (10) = (8) [R2 + R2 – 2v3 r] =) areV3b=0 so

ELL.LU let V2 = (i) be an eigen rector corresponding to 12 = 10 Then, (A-101) V2 = 0 13. (2)()-(3) >(v 3)(5)=68) [RPGSM * 6 oy)(945) [R = 82-203 Ri] 5) c-Vad = 0 do, vz = (m2). = c = 13 d. - : The matrix P that orthogonally diagonalizes A is given by - p=(V. Vz) 1. 1 U3 , / 23 2V3 - 2 so, 2V 15 \ I / 2 / 3 o papa / 203/15 12/15 Y5 116 QV3\/ 202 val. 2/15/lev3 7 ) 2) Baru BULSLLLLLL = D, diagonal form of a with eigen values as diagonal elements. Thus, p'ap=0, 0 = (0) - X-