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Question Number 141152 by ZiYangLee last updated on 16/May/21

$$\left(\begin{array}{cc}3& 5\\ 2& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\mathrm{and}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)$$$$\mathrm{are}\mathrm{true}\mathrm{for}\mathrm{any}\mathrm{values}\mathrm{of}x,y,x^{\prime },y^{\prime },$$$$\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}a,b,c,d.$$

Answered by iloveisrael last updated on 16/May/21

$$\Rightarrow \left(\begin{array}{c}35\\ 24\end{array}\right)\left(\begin{array}{c}ab\\ cd\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$$$$\Rightarrow \left(\begin{array}{c}ab\\ cd\end{array}\right)={\left(\begin{array}{c}35\\ 24\end{array}\right)}^{-1}=\frac{1}{2}\left(\begin{array}{c}4-5\\ -23\end{array}\right)$$$$=\left(\begin{array}{c}2-5/2\\ -13/2\end{array}\right)$$

Commented by ZiYangLee last updated on 16/May/21

$$\mathrm{Can}\mathrm{i}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{first}\mathrm{row}?$$

Commented by iloveisrael last updated on 16/May/21

$$bysubstitution$$

Answered by mr W last updated on 16/May/21

$$\left(\begin{array}{cc}3& 5\\ 2& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)asgiven$$$$\Rightarrow \left(\begin{array}{c}x\\ y\end{array}\right)={\left(\begin{array}{cc}3& 5\\ 2& 4\end{array}\right)}^{-1}\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$$$$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)asgiven$$$$\Rightarrow \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)={\left(\begin{array}{cc}3& 5\\ 2& 4\end{array}\right)}^{-1}=\left(\begin{array}{cc}2& -2.5\\ -1& 1.5\end{array}\right)$$

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