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Question Number 108059 by john santu last updated on 14/Aug/20

   ((♥JS♥)/(°js°))  Given a matrix A= (((   3       2)),((−5   −4)) )  and A^2 +♭A−2I=0 where ♭ is a  constant , I= (((1   0)),((0   1)) ). If B =   (((−3♭      2)),((   5♭    −1)) ) , then A^(−1) B =

$$\:\:\:\frac{\heartsuit{JS}\heartsuit}{°{js}°} \\ $$$${Given}\:{a}\:{matrix}\:{A}=\begin{pmatrix}{\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{5}\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$${and}\:{A}^{\mathrm{2}} +\flat{A}−\mathrm{2}{I}=\mathrm{0}\:{where}\:\flat\:{is}\:{a} \\ $$$${constant}\:,\:{I}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}.\:{If}\:{B}\:= \\ $$$$\begin{pmatrix}{−\mathrm{3}\flat\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\mathrm{5}\flat\:\:\:\:−\mathrm{1}}\end{pmatrix}\:,\:{then}\:{A}^{−\mathrm{1}} {B}\:=\: \\ $$

Answered by bemath last updated on 14/Aug/20

   ((⋋BeMath⋌)/⋎)  since ∣A∣ = −12+10=−2=−2∣I∣  then ♭ = −trace (A)=1 , so the equation becomes  A^2 +A=2I →2I=A(A+I)  by Cayley−Hamilton theorem   2A^(−1) =A+I ; A^(−1) =(1/2)(A+I)  A^(−1) =(1/2) (((    4       2)),((−5    −3)) ) and B= ((( −3      2)),((     5   −1)) )  therefore A^(−1) B=(1/2) (((−2        6)),((  0     −7)) )                 =  (((−1          3)),((   0      −(7/2))) )

$$\:\:\:\frac{\leftthreetimes{BeMath}\rightthreetimes}{\curlyvee} \\ $$$${since}\:\mid{A}\mid\:=\:−\mathrm{12}+\mathrm{10}=−\mathrm{2}=−\mathrm{2}\mid{I}\mid \\ $$$${then}\:\flat\:=\:−{trace}\:\left({A}\right)=\mathrm{1}\:,\:{so}\:{the}\:{equation}\:{becomes} \\ $$$${A}^{\mathrm{2}} +{A}=\mathrm{2}{I}\:\rightarrow\mathrm{2}{I}={A}\left({A}+{I}\right) \\ $$$${by}\:{Cayley}−{Hamilton}\:{theorem}\: \\ $$$$\mathrm{2}{A}^{−\mathrm{1}} ={A}+{I}\:;\:{A}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{I}\right) \\ $$$${A}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{5}\:\:\:\:−\mathrm{3}}\end{pmatrix}\:{and}\:{B}=\begin{pmatrix}{\:−\mathrm{3}\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\:\mathrm{5}\:\:\:−\mathrm{1}}\end{pmatrix} \\ $$$${therefore}\:{A}^{−\mathrm{1}} {B}=\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{6}}\\{\:\:\mathrm{0}\:\:\:\:\:−\mathrm{7}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:−\frac{\mathrm{7}}{\mathrm{2}}}\end{pmatrix} \\ $$

Commented by bemath last updated on 14/Aug/20

santuyy

$${santuyy} \\ $$

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