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Question Number 57805 by pete last updated on 12/Apr/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{y}=\mathrm{3x}+\mathrm{1}\:\mathrm{under}\:\mathrm{the} \\ $$$$\mathrm{mapping}\:\begin{pmatrix}{\mathrm{2}\:\:\:\mathrm{3}}\\{\mathrm{1}\:\:\:\mathrm{2}}\end{pmatrix}. \\ $$
Commented by pete last updated on 12/Apr/19
$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 12/Apr/19
$${point}\:\mathrm{1}\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\Rightarrow{image}\:\left(\mathrm{3},\mathrm{2}\right) \\ $$$${point}\:\mathrm{2}\:\left(−\frac{\mathrm{1}}{\mathrm{3}},\mathrm{0}\right) \\ $$$$\Rightarrow{image}\:\left(−\frac{\mathrm{2}}{\mathrm{3}},−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${eqn}.\:{of}\:{image}: \\ $$$$\frac{{y}+\frac{\mathrm{1}}{\mathrm{3}}}{{x}+\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{3}{y}+\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{11}} \\ $$$$\Rightarrow\mathrm{11}{y}=\mathrm{7}{x}+\mathrm{1} \\ $$
Answered by mr W last updated on 13/Apr/19
$${let}\:{image}\:{of}\:{point}\:\left({x},{y}\right)\:{be}\:\left({u},{v}\right), \\ $$$$\begin{pmatrix}{{u}}\\{{v}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{2}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}&{−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{2}}\end{pmatrix}\begin{pmatrix}{{u}}\\{{v}}\end{pmatrix} \\ $$$$\Rightarrow{x}=\mathrm{2}{u}−\mathrm{3}{v} \\ $$$$\Rightarrow{y}=−{u}+\mathrm{2}{v} \\ $$$$\Rightarrow−{u}+\mathrm{2}{v}=\mathrm{3}\left(\mathrm{2}{u}−\mathrm{3}{v}\right)+\mathrm{1} \\ $$$$\Rightarrow\mathrm{11}{v}=\mathrm{7}{u}+\mathrm{1} \\ $$$${i}.{e}.\:{image}\:{is}\:\mathrm{11}{y}=\mathrm{7}{x}+\mathrm{1} \\ $$