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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)) )

$$\lambda^{\mathrm{3}} −\left(\mathrm{trace}\:\mathrm{A}\right)\lambda^{\mathrm{2}} +\:\begin{pmatrix}{\mathrm{minor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{terms}}\\{\mathrm{on}\:\mathrm{the}\:\mathrm{leading}\:\mathrm{diag}\:\mathrm{A}\:}\end{pmatrix}\lambda−\mathrm{det}\left(\mathrm{A}\right)=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} +\left(\begin{vmatrix}{\mathrm{1}\:\:\mathrm{2}}\\{\mathrm{2}\:\:\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\mathrm{1}}\end{vmatrix}\right)\lambda−\mathrm{3}\:=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} +\lambda−\mathrm{3}\:=\:\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{Caley}−\mathrm{Hamilton}\:\mathrm{teorem} \\ $$$$\mathrm{A}^{\mathrm{3}} −\mathrm{3A}^{\mathrm{2}} +\mathrm{A}−\mathrm{3I}=\mathrm{0} \\ $$$$\mathrm{multiply}\:\mathrm{by}\:\mathrm{A}^{−\mathrm{1}} \\ $$$$ \:\mathrm{A}^{\mathrm{2}} −\mathrm{3A}+\mathrm{I}−\mathrm{3A}^{−\mathrm{1}} =\mathrm{0} \\ $$$$ \mathrm{3A}^{−\mathrm{1}} \:=\:\mathrm{A}^{\mathrm{2}} −\mathrm{3A}+\mathrm{I} \\ $$$$ \mathrm{3A}^{−\mathrm{1}} =\:\mathrm{A}\left(\mathrm{A}−\mathrm{3I}\right)+\mathrm{I} \\ $$$$ \mathrm{3A}^{−\mathrm{1}} =\:\begin{pmatrix}{−\mathrm{4}\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\mathrm{4}}\\{\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$ \mathrm{3A}^{−\mathrm{1}} =\:\begin{pmatrix}{−\mathrm{3}\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\mathrm{4}}\\{\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{3}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$ \mathrm{A}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{−\mathrm{3}\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\mathrm{4}}\\{\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$

Commented by BHOOPENDRA last updated on 27/Mar/21

thanks sir

$${thanks}\:{sir} \\ $$

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