Let \({e_1} = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\), \({e_2} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\), \({y_1} = \left[ {\begin{array}{*{20}{c}}2\\5\end{array}} \right]\) and \({y_2} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\6\end{array}} \right]\) and let \(T:{\mathbb{R}^2} \to {\mathbb{R}^2}\) be a linear transformation that maps \({e_1}\) into \({y_1}\) and maps \({e_2}\) into \({y_2}\). Find the images of \(\left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\).

Solve the equation \(x = \left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\).

\(\begin{aligned}x &= \left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}5\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0\\{ - 3}\end{array}} \right]\\ &= 5\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\\ &= 5{e_1} - 3{e_2}\end{aligned}\)

\(\begin{aligned}T\left( x \right) &= T\left( {5{e_1} - 3{e_2}} \right)\\ &= 5T\left( {{e_1}} \right) - 3T\left( {{e_2}} \right)\\ &= 5{y_1} - 3{y_2}\\ &= 5\left[ {\begin{array}{*{20}{c}}2\\5\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}{ - 1}\\6\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{13}\\7\end{array}} \right]\end{aligned}\)

\(x\)can be expressed as follows:

\(\begin{aligned}x &= \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\\ &= {x_1}\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\\ &= {x_1}{e_1} + {x_2}{e_2}\end{aligned}\)

\(\begin{aligned}{c}T\left( x \right) &= T\left( {{x_1}{e_1} + {x_2}{e_2}} \right)\\ &= {x_1}T\left( {{e_1}} \right) + {x_2}T\left( {{e_2}} \right)\\ &= {x_1}\left[ {\begin{array}{*{20}{c}}2\\5\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 1}\\6\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{2{x_1} - {x_2}}\\{5{x_1} + 6{x_2}}\end{array}} \right]\end{aligned}\)

So, the image of \(\left[ {\begin{array}{*{20}{c}}5\\{ - 3}\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}{13}\\7\end{array}} \right]\) and the image of \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}{2{x_1} - {x_2}}\\{5{x_1} + 6{x_2}}\end{array}} \right]\).