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Question-118896




Question Number 118896 by bobhans last updated on 20/Oct/20
Answered by benjo_mathlover last updated on 20/Oct/20
 → (((2     b)),((b      2)) )  ((x),(y) ) =  (((ac^2 +c)),((  c−1)) )  →  ((x),(y) ) = (1/(4−b^2 ))  (((2   −b)),((−b   2)) )  (((ac^2 +c)),((   c−1)) )  → ((x),(y) ) = (1/(4−b^2 ))  (((2ac^2 +2c−bc+b)),((−abc^2 −bc+2c−2)) )
$$\:\rightarrow\begin{pmatrix}{\mathrm{2}\:\:\:\:\:{b}}\\{{b}\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{{ac}^{\mathrm{2}} +{c}}\\{\:\:{c}−\mathrm{1}}\end{pmatrix} \\ $$$$\rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{4}−{b}^{\mathrm{2}} }\:\begin{pmatrix}{\mathrm{2}\:\:\:−{b}}\\{−{b}\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{{ac}^{\mathrm{2}} +{c}}\\{\:\:\:{c}−\mathrm{1}}\end{pmatrix} \\ $$$$\rightarrow\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{4}−{b}^{\mathrm{2}} }\:\begin{pmatrix}{\mathrm{2}{ac}^{\mathrm{2}} +\mathrm{2}{c}−{bc}+{b}}\\{−{abc}^{\mathrm{2}} −{bc}+\mathrm{2}{c}−\mathrm{2}}\end{pmatrix} \\ $$$$ \\ $$

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