Question

DETAILS LARLINALG87.1.006. Verify that 2, is an eigenvalue of A and that x, is a corresponding eigenvector. 5 -1 2 an = 5, xn = 1, 0, 0) A =10 3 11; Ax = 3,x; = 1, 2, 0) 0 0 4 .[ (1 ,1 ,1-) = ;x,4 = ܨܬ 1 AX, ܐܐ 5 -1 2 0 3 1 0 0 4 ܐܐ ]. 0 ܕܫܢ AX, - 5 -1 2 0 3 1 0 0 4 2 =_32 ܕX,2 AX3 5 -1_2 0 3 1 0 0 4 - [ - ܕܠ ܐܨܬ

Answer 1

solution Let A be a matrix then an a non zeno vector be A u is sald elger vector if there exist scalar d such that AV = du (d scalar) then elgen value corresponding elgen vector to verity that us A 2 X₂ is eigenvalue oq corresponding elgenuector sal 2 (1,0,0) AL do 25 O 3 1 da23 x 22 (1,2,0) X3 = (1, 1, 1) da - 4 5 AXI BE BE 5 -1 1 Multiply

than AX, 5xi so for X, [6] there exist a scalan dia 5 which implies x, mis eigenwecon A connesponding to eigenvalue 4,- 5 2 AX2 3 JE de X2 2 3 6 5-1 2 3 AX22 0 3 1 JB2 do AX22 3 X2 so for X2 2 [!] there exist a scalar da 2 3 which implies x2 is corresponding to A eigenvector eigenvalue da 2 3

-4 5 - 1 2 AX z ² 3 JE d 3 X3 5-12 AX₂ E] (3) 0 3 1 "El 04 • AXz2 4X₂ ( there exist a so for *32 scalar d3=4 which implies X3 is ergen a vector от А corresponding to eigen value da = 4