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 4βˆ’Ξ» βˆ’2 2 p(Ξ») = det βˆ’2 βˆ’1 βˆ’ Ξ» βˆ’3 2 βˆ’3 βˆ’1 βˆ’ Ξ» = (4 βˆ’ Ξ») det + (2) det βˆ’1 βˆ’ Ξ» βˆ’3 βˆ’3 βˆ’1 βˆ’ Ξ» βˆ’ (βˆ’2) det βˆ’2 βˆ’ Ξ» βˆ’1 βˆ’ Ξ» 2 βˆ’3 βˆ’2 βˆ’3 2 βˆ’1 βˆ’ Ξ» = (4 βˆ’ Ξ»){(Ξ» + 1)2 βˆ’ 9} + 2{2(1 + Ξ») + 6} + 2{6 + 2(1 + Ξ»)} = (4 βˆ’ Ξ»){(Ξ» + 4)(Ξ» βˆ’ 2)} + 2{2Ξ» + 8} + 2{2Ξ» + 8} = (4 βˆ’ Ξ»)(Ξ» + 4)(Ξ» βˆ’ 2) + 8(Ξ» + 4) = (Ξ» + 4){(4 βˆ’ Ξ»)(Ξ» βˆ’ 2) + 8} = (Ξ» + 4){βˆ’Ξ»2 + 6Ξ»} = βˆ’Ξ»(Ξ» + 4)(Ξ» βˆ’ 6). The eigenvalues of [Q] are Ξ»1 = 0, Ξ»2 = βˆ’4, Ξ»3 = 6. To find the eigenvectors for Ξ» = 0, we need 4v1 βˆ’2v2 +2v3 = 0 βˆ’2v1 βˆ’v2 βˆ’3v3 = 0 2v1 βˆ’3v2 βˆ’v3 = 0. The sum of the fist and second equations is the third, so we drop the last equation; dividing the first by 2, we have 2 v1 βˆ’ v2 + v3 = 0 βˆ’2v1 βˆ’v2 βˆ’3v3 = 0. The first of these two equations gives v2 = 2v1 + v3 and substituting this into the second gives βˆ’4v1 βˆ’ 4v2 = 0 ...
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