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Question Number 114554 by bemath last updated on 19/Sep/20

Given a matrix A =  (((a   b)),((c   d)) ) satisfies  the equation A^2 +λA+7I = 0  where I= (((1   0)),((0   1)) ) . Find the value of   trace of A

$${Given}\:{a}\:{matrix}\:{A}\:=\:\begin{pmatrix}{{a}\:\:\:{b}}\\{{c}\:\:\:{d}}\end{pmatrix}\:{satisfies} \\ $$$${the}\:{equation}\:{A}^{\mathrm{2}} +\lambda{A}+\mathrm{7}{I}\:=\:\mathrm{0} \\ $$$${where}\:{I}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\:.\:{Find}\:{the}\:{value}\:{of}\: \\ $$$${trace}\:{of}\:{A}\: \\ $$

Answered by bobhans last updated on 19/Sep/20

A^2 = (((a    b)),((c    d)) ) (((a    b)),((c     d)) )= (((a^2 +bc    ab+bd)),((ac+cd   bc+d^2 )) )  λA= (((aλ    bλ)),((cλ    dλ)) ) ⇒A^2 +λA= (((−7    0)),((   0   −7)) )  ⇒ (((a^2 +bc+λa    ab+bd+bλ)),((ac+cd+cλ   bc+d^2 +dλ)) )= (((−7      0)),((   0    −7)) )  → c(a+d+λ) = 0 ; a+d=−λ  →b(a+d+λ)=0 ; a+d =−λ  ⇒a^2 +bc+λa=−7  ⇒a^2 +bc+a(−a−d)=−7 ; bc−ad=−7  ad−bc = 7 ; det(A)=7  trace (A)=−λ

$${A}^{\mathrm{2}} =\begin{pmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:{d}}\end{pmatrix}\begin{pmatrix}{{a}\:\:\:\:{b}}\\{{c}\:\:\:\:\:{d}}\end{pmatrix}=\begin{pmatrix}{{a}^{\mathrm{2}} +{bc}\:\:\:\:{ab}+{bd}}\\{{ac}+{cd}\:\:\:{bc}+{d}^{\mathrm{2}} }\end{pmatrix} \\ $$$$\lambda{A}=\begin{pmatrix}{{a}\lambda\:\:\:\:{b}\lambda}\\{{c}\lambda\:\:\:\:{d}\lambda}\end{pmatrix}\:\Rightarrow{A}^{\mathrm{2}} +\lambda{A}=\begin{pmatrix}{−\mathrm{7}\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{0}\:\:\:−\mathrm{7}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{{a}^{\mathrm{2}} +{bc}+\lambda{a}\:\:\:\:{ab}+{bd}+{b}\lambda}\\{{ac}+{cd}+{c}\lambda\:\:\:{bc}+{d}^{\mathrm{2}} +{d}\lambda}\end{pmatrix}=\begin{pmatrix}{−\mathrm{7}\:\:\:\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{0}\:\:\:\:−\mathrm{7}}\end{pmatrix} \\ $$$$\rightarrow\:{c}\left({a}+{d}+\lambda\right)\:=\:\mathrm{0}\:;\:{a}+{d}=−\lambda \\ $$$$\rightarrow{b}\left({a}+{d}+\lambda\right)=\mathrm{0}\:;\:{a}+{d}\:=−\lambda \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{bc}+\lambda{a}=−\mathrm{7} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{bc}+{a}\left(−{a}−{d}\right)=−\mathrm{7}\:;\:{bc}−{ad}=−\mathrm{7} \\ $$$${ad}−{bc}\:=\:\mathrm{7}\:;\:{det}\left({A}\right)=\mathrm{7} \\ $$$${trace}\:\left({A}\right)=−\lambda \\ $$

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