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Question-212844

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Question Number 212844 by efronzo1 last updated on 25/Oct/24
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$$\:\:\:\cancel{\downharpoonleft}\underline{\:} \\ $$
Commented by AlagaIbile last updated on 29/Oct/24
det(S^2 ) = det^2 (S) = 36 det(S) = ± 6 By C-H Theorem: S^2 − tr(S) S + (±6) =0 Introduce trace to d eqn tr(S^2 )−tr^2 (S)+2(±6)=0 13 − tr^2 (S) ± 12 = 0 tr(S) = ± 5, ± 1 For trace(S) = ± 5 ∴ ±5S = S^2 + 6I_2 S = ± [(4,(-1)),(2,1) ] For trace(S) = ±1 ±S = S^2 − 6I_2 S = ± [(8,(-5)),((10),(-7)) ]
$$\:{det}\left({S}^{\mathrm{2}} \right)\:=\:{det}^{\mathrm{2}} \left({S}\right)\:=\:\mathrm{36} \\ $$$$\:{det}\left({S}\right)\:=\:\pm\:\mathrm{6} \\ $$$$\:{By}\:{C}-{H}\:{Theorem}: \\ $$$$\:\boldsymbol{{S}}^{\mathrm{2}} \:−\:{tr}\left(\boldsymbol{{S}}\right)\:\boldsymbol{{S}}\:+\:\left(\pm\mathrm{6}\right)\:=\mathrm{0} \\ $$$$\:{Introduce}\:{trace}\:{to}\:{d}\:{eqn} \\ $$$${tr}\left({S}^{\mathrm{2}} \right)−{tr}^{\mathrm{2}} \left({S}\right)+\mathrm{2}\left(\pm\mathrm{6}\right)=\mathrm{0} \\ $$$$\:\mathrm{13}\:−\:{tr}^{\mathrm{2}} \left({S}\right)\:\pm\:\mathrm{12}\:=\:\mathrm{0} \\ $$$$\:{tr}\left(\boldsymbol{{S}}\right)\:=\:\pm\:\mathrm{5},\:\pm\:\mathrm{1} \\ $$$$\:{For}\:{trace}\left(\boldsymbol{{S}}\right)\:=\:\pm\:\mathrm{5} \\ $$$$\:\therefore\:\:\pm\mathrm{5}\boldsymbol{{S}}\:=\:\boldsymbol{{S}}^{\mathrm{2}} \:+\:\mathrm{6}\boldsymbol{{I}}_{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{{S}}\:=\:\pm\:\begin{bmatrix}{\mathrm{4}}&{-\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\:{For}\:{trace}\left(\boldsymbol{{S}}\right)\:=\:\pm\mathrm{1} \\ $$$$\:\pm\boldsymbol{{S}}\:=\:\boldsymbol{{S}}^{\mathrm{2}} \:−\:\mathrm{6}\boldsymbol{{I}}_{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{{S}}\:=\:\pm\:\begin{bmatrix}{\mathrm{8}}&{-\mathrm{5}}\\{\mathrm{10}}&{-\mathrm{7}}\end{bmatrix} \\ $$
Answered by Rasheed.Sindhi last updated on 30/Oct/24
let S=± ((a,b),(c,d) ) ; assuming a,b,c,d∈Z S^2 =(± ((a,b),(c,d) ))^2 = (((a^2 +bc),(ab+bd)),((ca+cd),(bc+d^2 )) )= (((14 −5)),((10 −1)) ) a^2 +bc=14...i , b(a+d)=−5...ii c(a+d)=10...iii , bc+d^2 =−1...iv i − iv:a^2 −d^2 =15 { (((a−d)(a+d)=3×5⇒ { ((a−d=3)),((a+d=5)) :}⇒a=4,d=1)),(((a−d)(a+d)=5×3⇒ { ((a−d=5)),((a+d=3)) :}⇒a=4,d=−1)),(((a−d)(a+d)=1×15⇒ { ((a−d=1)),((a+d=15)) :}⇒a=8,d=7)),(((a−d)(a+d)=15×1⇒ { ((a−d=15)),((a+d=1)) :}⇒a=8,d=−7)) :} ii⇒ b=((−5)/(a+d)) iii⇒c=((10)/(a+d)) a=4,d=1⇒b=−1,c=2 a=4,d=−1⇒b=((−5)/3)∉Z , c=((10)/3)∉Z Rejected a=8,d=7⇒b=−(1/3)∉Z,c=((10)/(15))=(2/3)∉Z Rejected a=8,d=−7⇒b=−5,c=10 S=± ((4,(−1)),(2,( 1)) ) ,± ((8,(−5)),((10),(−7)) )
$$ \\ $$$${let}\:\mathrm{S}=\pm\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\:;\:{assuming}\:{a},{b},{c},{d}\in\mathbb{Z} \\ $$$$\mathrm{S}^{\mathrm{2}} =\left(\pm\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\right)^{\mathrm{2}} =\begin{pmatrix}{{a}^{\mathrm{2}} +{bc}}&{{ab}+{bd}}\\{{ca}+{cd}}&{{bc}+{d}^{\mathrm{2}} }\end{pmatrix}=\:\begin{pmatrix}{\mathrm{14}\:\:\:\:\:\:−\mathrm{5}}\\{\mathrm{10}\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\: \\ $$$${a}^{\mathrm{2}} +{bc}=\mathrm{14}…{i}\:\:,\:{b}\left({a}+{d}\right)=−\mathrm{5}…{ii} \\ $$$${c}\left({a}+{d}\right)=\mathrm{10}…{iii}\:,\:{bc}+{d}^{\mathrm{2}} =−\mathrm{1}…{iv} \\ $$$${i}\:−\:{iv}:{a}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{15} \\ $$$$\begin{cases}{\left({a}−{d}\right)\left({a}+{d}\right)=\mathrm{3}×\mathrm{5}\Rightarrow\begin{cases}{{a}−{d}=\mathrm{3}}\\{{a}+{d}=\mathrm{5}}\end{cases}\Rightarrow{a}=\mathrm{4},{d}=\mathrm{1}}\\{\left({a}−{d}\right)\left({a}+{d}\right)=\mathrm{5}×\mathrm{3}\Rightarrow\begin{cases}{{a}−{d}=\mathrm{5}}\\{{a}+{d}=\mathrm{3}}\end{cases}\Rightarrow{a}=\mathrm{4},{d}=−\mathrm{1}}\\{\left({a}−{d}\right)\left({a}+{d}\right)=\mathrm{1}×\mathrm{15}\Rightarrow\begin{cases}{{a}−{d}=\mathrm{1}}\\{{a}+{d}=\mathrm{15}}\end{cases}\Rightarrow{a}=\mathrm{8},{d}=\mathrm{7}}\\{\left({a}−{d}\right)\left({a}+{d}\right)=\mathrm{15}×\mathrm{1}\Rightarrow\begin{cases}{{a}−{d}=\mathrm{15}}\\{{a}+{d}=\mathrm{1}}\end{cases}\Rightarrow{a}=\mathrm{8},{d}=−\mathrm{7}}\end{cases} \\ $$$${ii}\Rightarrow\:{b}=\frac{−\mathrm{5}}{{a}+{d}}\:\:\:\:\:{iii}\Rightarrow{c}=\frac{\mathrm{10}}{{a}+{d}} \\ $$$$\: \\ $$$${a}=\mathrm{4},{d}=\mathrm{1}\Rightarrow{b}=−\mathrm{1},{c}=\mathrm{2} \\ $$$${a}=\mathrm{4},{d}=−\mathrm{1}\Rightarrow{b}=\frac{−\mathrm{5}}{\mathrm{3}}\notin\mathbb{Z}\:,\:{c}=\frac{\mathrm{10}}{\mathrm{3}}\notin\mathbb{Z}\:{Rejected}\: \\ $$$${a}=\mathrm{8},{d}=\mathrm{7}\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{3}}\notin\mathbb{Z},{c}=\frac{\mathrm{10}}{\mathrm{15}}=\frac{\mathrm{2}}{\mathrm{3}}\notin\mathbb{Z}\:{Rejected} \\ $$$${a}=\mathrm{8},{d}=−\mathrm{7}\Rightarrow{b}=−\mathrm{5},{c}=\mathrm{10} \\ $$$$\: \\ $$$$\mathrm{S}=\pm\begin{pmatrix}{\mathrm{4}}&{−\mathrm{1}}\\{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:,\pm\begin{pmatrix}{\mathrm{8}}&{−\mathrm{5}}\\{\mathrm{10}}&{−\mathrm{7}}\end{pmatrix}\: \\ $$
Commented by Ghisom last updated on 30/Oct/24
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Answered by Rasheed.Sindhi last updated on 26/Oct/24
let S= ((a,b),(c,d) ) S^2 = ((a,b),(c,d) )^2 = (((a^2 +bc),(ab+bd)),((ca+cd),(bc+d^2 )) )= (((14 −5)),((10 −1)) ) a^2 +bc=14 , b(a+d)=−5 c(a+d)=10 , bc+d^2 =−1 a+d=−(5/b)=((10)/c)⇒c=−2b bc=14−a^2 =−1−d^2 ⇒a^2 −d^2 =15 ⇒(a−d)(a+d)=3×5 let a−d=3 a+d=5⇒a=5−d S= (((5−d),b),((−2b),d) ) b=−1, d=1 One of answers: S= ((4,(−1)),(2,( 1)) ) Verification: S^2 = ((4,(−1)),(2,( 1)) ) ((4,(−1)),(2,( 1)) )= (((16−2),(−4−1)),((8+2),(−2+1)) )= (((14),(−5)),((10),(−1)) )
$${let}\:\mathrm{S}=\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix} \\ $$$$\mathrm{S}^{\mathrm{2}} =\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}^{\mathrm{2}} =\begin{pmatrix}{{a}^{\mathrm{2}} +{bc}}&{{ab}+{bd}}\\{{ca}+{cd}}&{{bc}+{d}^{\mathrm{2}} }\end{pmatrix}=\:\begin{pmatrix}{\mathrm{14}\:\:\:\:\:\:−\mathrm{5}}\\{\mathrm{10}\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\: \\ $$$${a}^{\mathrm{2}} +{bc}=\mathrm{14}\:\:,\:{b}\left({a}+{d}\right)=−\mathrm{5} \\ $$$${c}\left({a}+{d}\right)=\mathrm{10}\:,\:{bc}+{d}^{\mathrm{2}} =−\mathrm{1} \\ $$$${a}+{d}=−\frac{\mathrm{5}}{{b}}=\frac{\mathrm{10}}{{c}}\Rightarrow{c}=−\mathrm{2}{b} \\ $$$${bc}=\mathrm{14}−{a}^{\mathrm{2}} =−\mathrm{1}−{d}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{15} \\ $$$$\Rightarrow\left({a}−{d}\right)\left({a}+{d}\right)=\mathrm{3}×\mathrm{5} \\ $$$${let}\:\:{a}−{d}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:{a}+{d}=\mathrm{5}\Rightarrow{a}=\mathrm{5}−{d} \\ $$$$\mathrm{S}=\begin{pmatrix}{\mathrm{5}−{d}}&{{b}}\\{−\mathrm{2}{b}}&{{d}}\end{pmatrix} \\ $$$${b}=−\mathrm{1},\:{d}=\mathrm{1} \\ $$$${One}\:{of}\:{answers}: \\ $$$$\mathrm{S}=\begin{pmatrix}{\mathrm{4}}&{−\mathrm{1}}\\{\mathrm{2}}&{\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\mathcal{V}{erification}: \\ $$$$\mathrm{S}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{4}}&{−\mathrm{1}}\\{\mathrm{2}}&{\:\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{4}}&{−\mathrm{1}}\\{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{16}−\mathrm{2}}&{−\mathrm{4}−\mathrm{1}}\\{\mathrm{8}+\mathrm{2}}&{−\mathrm{2}+\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{14}}&{−\mathrm{5}}\\{\mathrm{10}}&{−\mathrm{1}}\end{pmatrix} \\ $$
Commented by Rasheed.Sindhi last updated on 26/Oct/24
ThanX sir! Pl see my next answer,where I covered all integral solutions.
$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$$${Pl}\:{see}\:{my}\:{next}\:{answer},{where} \\ $$$${I}\:{covered}\:{all}\:{integral}\:{solutions}. \\ $$
Commented by Ghisom last updated on 26/Oct/24
there are 4 solutions ± ((4,(−1)),(2,1) ) and ± ((8,(−5)),((10),(−7)) )
$$\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{solutions} \\ $$$$\pm\begin{pmatrix}{\mathrm{4}}&{−\mathrm{1}}\\{\mathrm{2}}&{\mathrm{1}}\end{pmatrix}\:\:\mathrm{and}\:\pm\begin{pmatrix}{\mathrm{8}}&{−\mathrm{5}}\\{\mathrm{10}}&{−\mathrm{7}}\end{pmatrix} \\ $$

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