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In Exercise 1-10, assume that \(T\) is a linear transformation. Find the standard matrix of \(T\).

\(T:{\mathbb{R}^3} \to {\mathbb{R}^2}\), rotates points (about the origin) through \(\frac{{3\pi }}{2}\)radians (counter-clockwise).

Short Answer

Expert verified

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

Step by step solution

01

Find the value of \(T\) using linear transformation

Using linear transformation,

\(\begin{aligned}{c}T &= T\left( {{x_1}{e_1} + {x_2}{e_2}} \right)\\ &= {x_1}T\left( {{e_1}} \right) + {x_2}T\left( {{e_2}} \right)\\ &= \left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_2}} \right)}\end{array}} \right]x\end{aligned}\)

02

Find the transformation for \(T\left( {{e_1}} \right)\) and \(T\left( {{e_2}} \right)\)

Transformation represents the rotation of\(\frac{{3\pi }}{2}\)radian about the origin (counterclockwise).

\(T\left( {{e_1}} \right) = - {e_2}\) and \(T\left( {{e_2}} \right) = {e_1}\).

03

Find the transformation for \(T\left( {{e_1}} \right)\) and \(T\left( {{e_2}} \right)\)

By the equation \(T = \left[ {\begin{array}{*{20}{c}}{T\left( {{e_1}} \right)}&{T\left( {{e_2}} \right)}\end{array}} \right]x\),

\(T = \left[ {\begin{array}{*{20}{c}}{ - {e_2}}&{{e_1}}\end{array}} \right]x\).

04

Find the standard matrix \(T\) for linear transformation

\({e_1} = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\)and\({e_2} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right]\).

In the equation \(T = Ax\), the matrix \(A\) is the matrix forlinear transformation \(T\).

By the equation \(T = \left[ {\begin{array}{*{20}{c}}{ - {e_2}}&{{e_1}}\end{array}} \right]x\),

\(T = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]x\).

Matrix \(A\) is \(\left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\).

So, the linear transformation matrix is \(\left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\).

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Most popular questions from this chapter

In Exercises 6, write a system of equations that is equivalent to the given vector equation.

6. \({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)

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