Question

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

\(T\left( {{x_1},{x_2}} \right) = \left( {2{x_2} - 3{x_1},{x_1} - 4{x_2},0,{x_2}} \right)\)

Short answer

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

Step-by-step solution

Express \(T\left( x \right)\) in the form of a matrix

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

\(T\left( x \right) = \left[ {\begin{array}{*{20}{c}}{2{x_2} - 3{x_1}}\\{{x_1} - 4{x_2}}\\0\\{{x_2}}\end{array}} \right]\)

Solve the equation \(T\left( x \right) = Ax\)

\(\left[ {\begin{array}{*{20}{c}}{2{x_2} - 3{x_1}}\\{{x_1} - 4{x_2}}\\0\\{{x_2}}\end{array}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\)

As \(\left[ x \right]\) has only two entries, matrix \(A\) will have two columns and four rows.

Compare the rows of the matrix

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

Compare the rows of the matrix

From the equation \(\left[ {\begin{array}{*{20}{c}}{2{x_2} - 3{x_1}}\\{{x_1} - 4{x_2}}\\0\\{{x_2}}\end{array}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\), the second row of matrix \(A\) is \(\left[ {\begin{array}{*{20}{c}}1&{ - 4}\end{array}} \right]\).

Compare the rows of the matrix

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

Compare the rows of the matrix

From the equation \(\left[ {\begin{array}{*{20}{c}}{2{x_2} - 3{x_1}}\\{{x_1} - 4{x_2}}\\0\\{{x_2}}\end{array}} \right] = \left[ A \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right]\), the third row of matrix \(A\) is \(\left[ {\begin{array}{*{20}{c}}0&1\end{array}} \right]\).

So, the matrix given in the equation is \(\left[ {\begin{array}{*{20}{c}}{ - 3}&2\\1&{ - 4}\\0&0\\0&1\end{array}} \right]\).