Let \(T:{\mathbb{R}^2} \to {\mathbb{R}^3}\) be a linear transformation such that \(T\left( {{x_1},{x_2}} \right) = \left( {{x_1} - 2{x_2}, - {x_1} + 3{x_2},3{x_1} - 2{x_2}} \right)\). Find \({\mathop{\rm x}\nolimits} \) such that \(T\left( x \right) = \left( { - 1,4,9} \right)\).

It is given that \(T\left( x \right) = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\\9\end{array}} \right]\).

Write the linear transformation into the standard matrix of \(T\) by inspection.

\[\begin{array}{c}T\left( x \right) = \left[ {\begin{array}{*{20}{c}}{{x_1} - 2{x_2}}\\{ - {x_1} + 3{x_2}}\\{3{x_1} - 2{x_2}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{ - 1}\\4\\9\end{array}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{ - 1}\\4\\9\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\{ - 1}&3\\3&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\end{array}\]

Write the standard matrix into an augmented matrix.

\[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 1}\\{ - 1}&3&4\\3&{ - 2}&9\end{array}} \right]\]

At row 2, multiply row 1 by 1 and add it to row 2.

\[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 1}\\0&1&3\\3&{ - 2}&9\end{array}} \right]\]

At row 3, multiply row 1 by \(3\)and subtract it from row 3.

\[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&{ - 1}\\0&1&3\\0&4&{12}\end{array}} \right]\]

At row 1, multiply row 2 by 2 and add it to row 1.

\[\left[ {\begin{array}{*{20}{c}}1&0&5\\0&1&3\\0&4&{12}\end{array}} \right]\]

At row 3, multiply row 2 by \(4\) and subtract it from row 3.

\[\left[ {\begin{array}{*{20}{c}}1&0&5\\0&1&3\\0&0&0\end{array}} \right]\]

Thus, the value of \(x\) is \(x = \left[ {\begin{array}{*{20}{c}}5\\3\end{array}} \right]\) such that \(T\left( x \right) = \left( { - 1,4,9} \right)\).