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Question-136885

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Question Number 136885 by BHOOPENDRA last updated on 27/Mar/21
Answered by bramlexs22 last updated on 27/Mar/21
λ^3 −(trace A)λ^2 + (((minor of the terms)),((on the leading diag A )) )λ−det(A)=0 λ^3 −3λ^2 +( determinant (((1 2)),((2 1)))+ determinant (((1 0)),((1 1)))+ determinant ((( 1 2)),((−1 1))))λ−3 =0 λ^3 −3λ^2 +λ−3 = 0 by Caley−Hamilton teorem A^3 −3A^2 +A−3I=0 multiply by A^(−1) A^2 −3A+I−3A^(−1) =0 3A^(−1) = A^2 −3A+I 3A^(−1) = A(A−3I)+I 3A^(−1) = (((−4 −2 4)),(( 3 0 −2)),((−3 0 2)) ) + (((1 0 0)),((0 1 0)),((0 0 1)) ) 3A^(−1) = (((−3 −2 4)),(( 3 1 −2)),((−3 0 3)) ) A^(−1) = (1/3) (((−3 −2 4)),(( 3 1 −2)),((−3 0 3)) )
λ3−(traceA)λ2+(minorofthetermsontheleadingdiagA)λ−det(A)=0λ3−3λ2+(|1221|+|1011|+|12−11|)λ−3=0λ3−3λ2+λ−3=0byCaley−HamiltonteoremA3−3A2+A−3I=0multiplybyA−1A2−3A+I−3A−1=03A−1=A2−3A+I3A−1=A(A−3I)+I3A−1=(−4−2430−2−302)+(100010001)3A−1=(−3−2431−2−303)A−1=13(−3−2431−2−303)
Commented by BHOOPENDRA last updated on 27/Mar/21
thanks sir
thankssir

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