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Question Number 122579 by bramlexs22 last updated on 18/Nov/20

Answered by liberty last updated on 18/Nov/20

(2a) Q symetric if Q = Q^t ⇔ (((5 6x+5y 1)),((4 6 5y^2 −2x)),((x^2 −8 3z p)) ) = (((5 4 x^2 −8)),((6x+5y 6 3z)),(( 1 5y^2 −2x p)) ) gives { ((6x+5y=1 ∧ x^2 −8=1 { ((x=3)),((x=−3)) :})),((y= { ((((1−6(3))/5)=−((17)/5))),((((1−6(−3))/5)=((19)/5))) :})) :} ⇔ z = ((5y^2 −2x)/3)= { (((5(−((17)/5))^2 −2.3)/3)),(((5(((19)/5))^2 −2.(−3))/3)) :}

$$\left(\mathrm{2a}\right)\:\mathrm{Q}\:\mathrm{symetric}\:\mathrm{if}\:\mathrm{Q}\:=\:\mathrm{Q}^{\mathrm{t}} \\ $$$$\Leftrightarrow\:\begin{pmatrix}{\mathrm{5}\:\:\:\:\:\mathrm{6x}+\mathrm{5y}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\mathrm{5y}^{\mathrm{2}} −\mathrm{2x}}\\{\mathrm{x}^{\mathrm{2}} −\mathrm{8}\:\:\:\mathrm{3z}\:\:\:\:\:\:\:\:\:\:\:\mathrm{p}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{8}}\\{\mathrm{6x}+\mathrm{5y}\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\mathrm{3z}}\\{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5y}^{\mathrm{2}} −\mathrm{2x}\:\:\:\mathrm{p}}\end{pmatrix} \\ $$$$\mathrm{gives}\:\begin{cases}{\mathrm{6x}+\mathrm{5y}=\mathrm{1}\:\wedge\:\mathrm{x}^{\mathrm{2}} −\mathrm{8}=\mathrm{1}\:\begin{cases}{\mathrm{x}=\mathrm{3}}\\{\mathrm{x}=−\mathrm{3}}\end{cases}}\\{\mathrm{y}=\begin{cases}{\frac{\mathrm{1}−\mathrm{6}\left(\mathrm{3}\right)}{\mathrm{5}}=−\frac{\mathrm{17}}{\mathrm{5}}}\\{\frac{\mathrm{1}−\mathrm{6}\left(−\mathrm{3}\right)}{\mathrm{5}}=\frac{\mathrm{19}}{\mathrm{5}}}\end{cases}}\end{cases} \\ $$$$\:\Leftrightarrow\:\mathrm{z}\:=\:\frac{\mathrm{5y}^{\mathrm{2}} −\mathrm{2x}}{\mathrm{3}}=\begin{cases}{\frac{\mathrm{5}\left(−\frac{\mathrm{17}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{2}.\mathrm{3}}{\mathrm{3}}}\\{\frac{\mathrm{5}\left(\frac{\mathrm{19}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{2}.\left(−\mathrm{3}\right)}{\mathrm{3}}}\end{cases} \\ $$

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