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

\(T:{\mathbb{R}^2} \to {\mathbb{R}^2}\), first performs a horizontal share that transforms \({e_2}\) into \({e_2} - 2{e_1}\) (leaving \({e_1}\) unchanged) and then reflects point through the line the line \({x_2} = - {x_1}\).

Usinglinear transformation,

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

For \({e_1}\), the horizontal shear leaves it unchanged.

\({e_1} \to {e_1}\)

When it is reflected through the line \({x_2} = - {x_1}\),

\({e_1} \to - {e_2}\)

For \({e_2}\), the horizontal shear transforms \({e_2}\) into \({e_2} - 2{e_1}\).

\({e_2} \to {e_2} - 2{e_1}\)

When it is reflected through the line \({x_2} = - {x_1}\) (the image would be the same as a linear combination of images of \({e_2}\) and \({e_1}\), i.e., \({e_2} \to - {e_1}\) and \({e_1} \to - {e_2}\)),

\({e_2} - 2{e_1} \to - {e_1} + 2{e_2}\).

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

\(\begin{aligned} T &= \left[ {\begin{array}{*{20}{c}}{ - {e_2}}&{ - {e_1}}\end{array} + 2{e_2}} \right]x\\ &= \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\{ - 1}&2\end{array}} \right]x\end{aligned}\)

By the equation \(T = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\{ - 1}&2\end{array}} \right]x\), the matrix \(A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\{ - 1}&2\end{array}} \right]\).

So, thelinear transformation matrix is \(\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\{ - 1}&2\end{array}} \right]\).