Question

3. ( Find all eigenvalues and eigenvectors of the matrix A= [ 5 | 3 -1] and show the eigen- 1 vectors are linearly independent.

Answer 1

sol? Azt 7 för Linding Eigen valuesCharacterstic polynomial is LA-HIL=0 > || 5-1-tio -o. - 3 lois *] 5-1 11 30 (5-2)(1-2). - 3X(-1) = 0 5-57-7+1²+3=0 ²-60+8=0 3²-4-28+8=0 eld-4) -2 (0-4)=0 > 14-2) (6-4)=0 = -2=0 or +420 * d=2 or d=4 Tlmiy values are a= 2 and da=4 / Now fond-Ligen vector for corresponding eyes velues. Elgen vestes for pred for eigen vector A-dI) v =0

| 2 (A- AI) AL) br /15 11 - - || 5 위 | 275-2 I Its [3 - // r 32 3 1 Now (A-I) 0=0 3거 예 3게 we have a homogeneous system of solve it by Gaussian elmination. linear egh we [ 3 - 1 1: Ris Rixto ㅋ R, - 3A " 낯l 이 L3 IT] 5101 키 개농장 라드 등라자 24 heherl 807 : " 5월 = g 건g but nal

for da = 4 Eyen vector can find as A-da I = 15-17- 5-4 3 -1 - 4 y 3 3. il ol A-da I = -11 3-31 M solve it Now for eigen vector (A-d ) v =0. so we have a homogeneous system of linear eghs by using heussian elimination! Il-110 Ron R2 R2 - 3RI 13- 30 n -lo7 Loo 101 » 21-Hq=0 = 2 - Hg = ng General soin, X = / 12 /= Hall 7 - put ng=1 T = (1 7 Required eigen vector for de su

The we show that Eigen welters are linearly independent vectors şurah in IR is unearly Independent if the only solution of Civita U2=0 is c.=(2=0 I c + C₂7 CitC2 Tv TC7 Toy II 2 Tito Ī I Teat we have a homogeneous system of eph T ₃ llo. Rim R,X3. ni 37o/ ㅠ 고이 Ra > R2 -RI I 3 T! 3107 R > R2 X - _ _ _ -2 R, R, - 3R2

urlolo7 - » ci=0 & (2=0 is the only solution of slegs ŞU,, We} are linearly independent weeters. 2