Chapter 6 - Linear Transformations - 6.1 Introduction to Linear Transformations - 6.1 Exercises - Page 301: 56

$T\left( {\left[ {\begin{array}{*{20}{c}} 1&3\\ { - 1}&4 \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}} {12}&{ - 1}\\ 7&4 \end{array}} \right]$

Observe that, $\left[ {\begin{array}{*{20}{c}} 1&3\\ { - 1}&4 \end{array}} \right] = 1\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&0 \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 0&1\\ 0&0 \end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right] + 4\left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]$ Since, $T$ is a linear transformation, $\begin{array}{l} T\left( {\left[ {\begin{array}{*{20}{c}} 1&3\\ { - 1}&4 \end{array}} \right]} \right)\\ = T\left( {1\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&0 \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 0&1\\ 0&0 \end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right] + 4\left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]} \right)\\ = 1T\left( {1\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&0 \end{array}} \right]} \right) + 3T\left( {3\left[ {\begin{array}{*{20}{c}} 0&1\\ 0&0 \end{array}} \right]} \right) - 1T\left( {\left[ {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right]} \right) + 4T\left( {\left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]} \right)\\ = 1\left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&2 \end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}} 0&2\\ 1&1 \end{array}} \right] - 1\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&1 \end{array}} \right] + 4\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 1&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {12}&{ - 1}\\ 7&4 \end{array}} \right] \end{array}$