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-8 -24 -12 (16 points) Let A= 0 4 0 6 12 10 (a) (4 points) Find the eigenvalues of A. (b) [6 points) For each eigenvalue of A
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Soint Given Matrox -24 -12 O 4 0 6 12 10 (@) for Eigen values (A-XIl=0 -24 -12 0 u-X 6 12 10- (+850) [cu-1) (10-1)] +24 [0]=ab) now for a - 4 Eigen vetor X = 3 22 3 (A-41) X=0 Egg BING R37223 + Rp ية -120, 2424 - 1222 = 0 x2 & 3 are free variable X -Similouly for 1= -2,5) (A+21) X=0 -24 24 0 6 12 (2 R₂ Rythy & R₂ R₂ +2 R2 -24 6 22 0 By comparing -68, -24% -12% 62 X = 0 X₂(C) for a = 4 for a = -2 2 a Characteristic Equation only linear factor so A is diagnoşable 0 D 0 12 0 0 ds O 0 0 0 T P where

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