Engineering Calculus Notes 392

Engineering Calculus Notes 392 - 380 CHAPTER 3 REAL-VALUED...

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Unformatted text preview: 380 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION with characteristic polynomial 4βΞ» β2 2 p(Ξ») = det β2 β1 β Ξ» β3 2 β3 β1 β Ξ» = (4 β Ξ») det + (2) det β1 β Ξ» β3 β3 β1 β Ξ» β (β2) det β2 β Ξ» β1 β Ξ» 2 β3 β2 β3 2 β1 β Ξ» = (4 β Ξ»){(Ξ» + 1)2 β 9} + 2{2(1 + Ξ») + 6} + 2{6 + 2(1 + Ξ»)} = (4 β Ξ»){(Ξ» + 4)(Ξ» β 2)} + 2{2Ξ» + 8} + 2{2Ξ» + 8} = (4 β Ξ»)(Ξ» + 4)(Ξ» β 2) + 8(Ξ» + 4) = (Ξ» + 4){(4 β Ξ»)(Ξ» β 2) + 8} = (Ξ» + 4){βΞ»2 + 6Ξ»} = βΞ»(Ξ» + 4)(Ξ» β 6). The eigenvalues of [Q] are Ξ»1 = 0, Ξ»2 = β4, Ξ»3 = 6. To ο¬nd the eigenvectors for Ξ» = 0, we need 4v1 β2v2 +2v3 = 0 β2v1 βv2 β3v3 = 0 2v1 β3v2 βv3 = 0. The sum of the ο¬st and second equations is the third, so we drop the last equation; dividing the ο¬rst by 2, we have 2 v1 β v2 + v3 = 0 β2v1 βv2 β3v3 = 0. The ο¬rst of these two equations gives v2 = 2v1 + v3 and substituting this into the second gives β4v1 β 4v2 = 0 ...
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