Question

Let \(T:{\mathbb{R}^2} \to {\mathbb{R}^2}\) be a linear transformation that maps \(u = \left[ {\begin{array}{*{20}{c}}5\\2\end{array}} \right]\) into \(\left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\) and maps \(v = \left[ {\begin{array}{*{20}{c}}1\\3\end{array}} \right]\) into \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right]\). Use the fact that \(T\) is linear to find the images under \(T\) of \(3u\), \(2v\) and \(3u + 2v\).

Short answer

\(T\left( {3u} \right) = \left[ {\begin{array}{*{20}{c}}6\\3\end{array}} \right]\), \(T\left( {2v} \right) = \left[ {\begin{array}{*{20}{c}}{ - 2}\\6\end{array}} \right]\), and \(T\left( {3u + 2v} \right) = \left[ {\begin{array}{*{20}{c}}4\\9\end{array}} \right]\)

Step-by-step solution

Finding the transformed coordinates

The transformed rectangular coordinate of \(u\) is:

\(T\left( u \right) = \left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\)

The transformed rectangular coordinate of \(v\) is:

\(T\left( v \right) = \left[ {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right]\)

Finding the image of rectangular coordinates

Calculate the value of \(T\left( {3u} \right)\) as follows:

\(\begin{aligned}{c}T\left( {3u} \right) &= 3T\left( u \right)\\ &= 3\left[ {\begin{array}{*{20}{c}}2\\1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}6\\3\end{array}} \right]\end{aligned}\)

Finding the image of the rectangular coordinates

Calculate the value of \(T\left( {2v} \right)\) as follows:

\(\begin{aligned}T\left( {2v} \right) &= 2T\left( v \right)\\ &= 2\left[ {\begin{array}{*{20}{c}}{ - 1}\\3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 2}\\6\end{array}} \right]\end{aligned}\)

Finding the image of the rectangular coordinates

Calculate the value of \(T\left( {3u + 2v} \right)\) as follows:

\(\begin{aligned} T\left( {3u + 2v} \right) &= T\left( {3u} \right) + T\left( {2v} \right)\\ &= \left[ {\begin{array}{*{20}{c}}6\\3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 2}\\6\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}4\\9\end{array}} \right]\end{aligned}\)

So, the values of the images are \(T\left( {3u} \right) = \left[ {\begin{array}{*{20}{c}}6\\3\end{array}} \right]\), \(T\left( {2v} \right) = \left[ {\begin{array}{*{20}{c}}{ - 2}\\6\end{array}} \right],\)and \(T\left( {3u + 2v} \right) = \left[ {\begin{array}{*{20}{c}}4\\9\end{array}} \right]\).