In Exercises 17-20, show that \(T\) is a linear transformation by finding a matrix that implements the mapping. Note that \({x_1}\), \({x_2}\),……… are not vectors but are enteries in vectors

\(T\left( {{x_1},{x_2},{x_3},{x_4}} \right) = 2{x_1} + 3{x_3} - 4{x_4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {T:{\mathbb{R}^4} \to \mathbb{R}} \right)\)

Write the linear transformation \(T\left( x \right)\).

\(T\left( x \right) = \left[ {2{x_1} + 3{x_3} - 4{x_4}} \right]\)

\(\left[ {2{x_1} + 3{x_3} - 4{x_4}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\)

As \(T\left( x \right)\) has only one entry, matrix \(A\) will have one row and \(x\) has four entries so matrix A will have four columns.

From the equation \(\left[ {2{x_1} + 3{x_3} - 4{x_4}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\), the first row of matrix \(A\) is \(\left[ {\begin{array}{*{20}{c}}2&0&3&{ - 4}\end{array}} \right]\).

So, the matrix given in the equation is \(\left[ {\begin{array}{*{20}{c}}2&0&3&{ - 4}\end{array}} \right]\).