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Question Number 109240 by bemath last updated on 22/Aug/20
((♭emath)/(•••••)) use cayley − hamilton theorem to calculate A^(−1) for A= (((1 2 2)),((1 2 −1)),((−1 1 4)) )
$$\:\:\frac{\flat{emath}}{\bullet\bullet\bullet\bullet\bullet} \\ $$$${use}\:{cayley}\:−\:{hamilton}\:{theorem} \\ $$$${to}\:{calculate}\:{A}^{−\mathrm{1}} \:{for}\:{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{2}\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\mathrm{1}\:\:\:\:\mathrm{4}}\end{pmatrix} \\ $$
Answered by bobhans last updated on 22/Aug/20
((♭o♭hans)/(≤≥≡•°#)) the characteristic equation λ^3 −(tr A)λ^2 + ((( minor )),((from diagonal)) ) λ−det(A)=0 λ^3 −9λ^2 + ( determinant (((2 −1)),((1 4)))+ determinant ((( 1 2)),((−1 4)))+ determinant (((1 2)),((1 2))))λ−det(A)=0 λ^3 −9λ^2 +15λ−9=0 By Cayley − Hamilton theorem A^3 −9A^2 +15A−9I=0 multiply both sides by A^(−1) ⇒A^2 −9A+15I−9A^(−1) =0 9A^(−1) =A^2 −9A+15I 9A^(−1) = (((1 8 8)),((4 5 −4)),((−4 4 13)) ) − (((9 18 18)),((9 18 −9)),((−9 9 36)) ) + (((15 0 0)),((0 15 0)),((0 0 15)) ) 9A^(−1) = ((( 7 −10 −10)),((−5 2 5)),(( 5 −5 −8)) ) ∴ A^(−1) = (1/9) ((( 7 −10 −10)),((−5 2 5)),(( 5 −5 −8)) )
$$\:\:\:\:\frac{\flat{o}\flat{hans}}{\leqslant\geqslant\equiv\bullet°#} \\ $$$${the}\:{characteristic}\:{equation}\: \\ $$$$\lambda^{\mathrm{3}} −\left({tr}\:{A}\right)\lambda^{\mathrm{2}} +\begin{pmatrix}{\:\:\:\:\:\:\:{minor}\:}\\{{from}\:{diagonal}}\end{pmatrix}\:\lambda−{det}\left({A}\right)=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{9}\lambda^{\mathrm{2}} +\:\left(\begin{vmatrix}{\mathrm{2}\:\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\mathrm{4}}\end{vmatrix}+\begin{vmatrix}{\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\mathrm{4}}\end{vmatrix}+\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\mathrm{2}}\end{vmatrix}\right)\lambda−{det}\left({A}\right)=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{15}\lambda−\mathrm{9}=\mathrm{0} \\ $$$${By}\:{Cayley}\:−\:{Hamilton}\:{theorem} \\ $$$${A}^{\mathrm{3}} −\mathrm{9}{A}^{\mathrm{2}} +\mathrm{15}{A}−\mathrm{9}{I}=\mathrm{0} \\ $$$${multiply}\:{both}\:{sides}\:{by}\:{A}^{−\mathrm{1}} \\ $$$$\Rightarrow{A}^{\mathrm{2}} −\mathrm{9}{A}+\mathrm{15}{I}−\mathrm{9}{A}^{−\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{9}{A}^{−\mathrm{1}} ={A}^{\mathrm{2}} −\mathrm{9}{A}+\mathrm{15}{I} \\ $$$$\mathrm{9}{A}^{−\mathrm{1}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\:\mathrm{8}}\\{\mathrm{4}\:\:\:\:\:\:\:\mathrm{5}\:\:−\mathrm{4}}\\{−\mathrm{4}\:\:\:\mathrm{4}\:\:\:\:\mathrm{13}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{9}\:\:\:\:\:\mathrm{18}\:\:\:\:\mathrm{18}}\\{\mathrm{9}\:\:\:\:\:\:\mathrm{18}\:\:\:\:−\mathrm{9}}\\{−\mathrm{9}\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{36}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{15}\:\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{15}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\mathrm{15}}\end{pmatrix} \\ $$$$\mathrm{9}{A}^{−\mathrm{1}} =\begin{pmatrix}{\:\mathrm{7}\:\:\:\:−\mathrm{10}\:\:\:\:\:−\mathrm{10}}\\{−\mathrm{5}\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{\:\:\mathrm{5}\:\:\:\:\:\:\:\:−\mathrm{5}\:\:\:\:\:\:\:\:−\mathrm{8}}\end{pmatrix} \\ $$$$\therefore\:{A}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{9}}\begin{pmatrix}{\:\:\:\mathrm{7}\:\:\:\:\:\:\:−\mathrm{10}\:\:\:\:\:\:\:−\mathrm{10}}\\{−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}}\\{\:\:\:\mathrm{5}\:\:\:\:\:\:\:\:\:−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{8}}\end{pmatrix} \\ $$
Commented by john santu last updated on 22/Aug/20

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