Question

0 0 Q2. Consider the matrix A 6 2 -5 0 1 (a) Find all eigenvalues of the matrix A. (7 pts) (b) Find all eigenvectors of the matrix A. (8 pts) (c) Do you think that the set of the eigenvectors of A is a basis for the vector space R*? (Justify your answer) (5 pts)

Answer 1

0 2. Az 6 2 -5 o -6 -5 0 1 (a) Characteristic equation A is 6-X 2 -5-X -5 O I- (6-1) { (-5 - 1) (1-1) - 0} = 0 (6x) (-5-x) (1-x) = 0 x = 6, 5, 1 Let ^2=-5, 3-1 X = 6, Eigen values of A are 6, -5, 1 rector. (b) For x1=6, be x= eigen O Ax = x,x → (A-X,I) x=0 - 11 where A-X, I= -5 O -5 equation and We have a homogeneous Gaussion system of linear Elimination, solve it by we -3 O 2. -11 О о о -60 R₂ HP, 0 o O -11 -6 ооо O -5 O -5 -5 -5 O o 1-5 -310 -3 1 1 R₂ R₂ R3 +5R, O O -55 -20 Na оо -20 o 0 - 1 -3 - ES Rg 1 - 11 So R + 1/ R₂ 유 0 - 11 оо O 0 O 2, +23 x = -xz - x₂ = - 3 x3

X = 03 Floo - 23 For ^2 = 5 Ax= x2x оо O 6 » (A - x₂) x=0 where A - XI 2. -5 0 6 equations. There is homogeneous system of linear 'Elimination We solve Gaussian it by 1 010 11 1 o 0-60 1 XR O 010 T 2 0 -6 0-60 R₂-28, R3+5R, O O O ооо 2 0 610 -5 6 6 5 1 - 0 00 R3-6R2 OO 1 - // xR2 O 0 0 0 O O O 0 O ox-o 0 x = = x2 for Az = 1 AX= x 3 x 5 2. - 5 0 -6 -6 where A - Az I = (A - X3 I)x=0 of linear equation, There is we solve homogeneous Gaussian system Elimination it by 0 - ооо -6-60 1 0 O 0 -6 -6 o 010 2 -6 -60 2 R2-2R, R.3 +5R, -5 o 0 0 O 0 O 0 -5 o 1 0 010 R2 1 O 1:2 00 x2 + x3=0 12--23

- 3 아 외3 so A of vectors eigen \ 8 1.1.]
This is your answer .