Question Number 108059 by john santu last updated on 14/Aug/20
$$\:\:\:\frac{\heartsuit{JS}\heartsuit}{°{js}°} \\ $$$${Given}\:{a}\:{matrix}\:{A}=\begin{pmatrix}{\:\:\:\mathrm{3}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{5}\:\:\:−\mathrm{4}}\end{pmatrix} \\ $$$${and}\:{A}^{\mathrm{2}} +\flat{A}−\mathrm{2}{I}=\mathrm{0}\:{where}\:\flat\:{is}\:{a} \\ $$$${constant}\:,\:{I}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}.\:{If}\:{B}\:= \\ $$$$\begin{pmatrix}{−\mathrm{3}\flat\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\mathrm{5}\flat\:\:\:\:−\mathrm{1}}\end{pmatrix}\:,\:{then}\:{A}^{−\mathrm{1}} {B}\:=\: \\ $$
Answered by bemath last updated on 14/Aug/20
$$\:\:\:\frac{\leftthreetimes{BeMath}\rightthreetimes}{\curlyvee} \\ $$$${since}\:\mid{A}\mid\:=\:−\mathrm{12}+\mathrm{10}=−\mathrm{2}=−\mathrm{2}\mid{I}\mid \\ $$$${then}\:\flat\:=\:−{trace}\:\left({A}\right)=\mathrm{1}\:,\:{so}\:{the}\:{equation}\:{becomes} \\ $$$${A}^{\mathrm{2}} +{A}=\mathrm{2}{I}\:\rightarrow\mathrm{2}{I}={A}\left({A}+{I}\right) \\ $$$${by}\:{Cayley}−{Hamilton}\:{theorem}\: \\ $$$$\mathrm{2}{A}^{−\mathrm{1}} ={A}+{I}\:;\:{A}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{I}\right) \\ $$$${A}^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{5}\:\:\:\:−\mathrm{3}}\end{pmatrix}\:{and}\:{B}=\begin{pmatrix}{\:−\mathrm{3}\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\:\mathrm{5}\:\:\:−\mathrm{1}}\end{pmatrix} \\ $$$${therefore}\:{A}^{−\mathrm{1}} {B}=\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{6}}\\{\:\:\mathrm{0}\:\:\:\:\:−\mathrm{7}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\\{\:\:\:\mathrm{0}\:\:\:\:\:\:−\frac{\mathrm{7}}{\mathrm{2}}}\end{pmatrix} \\ $$
Commented by bemath last updated on 14/Aug/20
$${santuyy} \\ $$