Let $T : \Bbb{R}^2 \rightarrow \Bbb{R}^2$ be the transformation that first performs a horizontal shear so that $e_2 \rightarrow e_2 + 2e_1$ (leaving $e_1$ unchanged) and then reflects the points through the line $y = x$
(a) Find the standard matrix $A$ for $T$
(b) Find the standard matrix for the inverse mapping directly by finding a transformation that first undoes the reflection through the line y=x and then undoes the horizontal shear. Show that this matrix is the same as the matrix you found for $A^{-1}$
I'm on (a) and have
$\underbrace{
\begin{bmatrix}
x_1 & y_1\\
x_2 & y_2\\
\end{bmatrix}
}_{M}$
$\xrightarrow{(2)e_1 + e_2 \rightarrow e_2}
\begin{bmatrix}
x_1 & y_1\\
2x_1+x_2 & 2y_1+y_2\\
\end{bmatrix}$
$\begin{bmatrix}
0 & 1\\
1 & 0\\
\end{bmatrix}
\begin{bmatrix}
x_1 & y_1\\
2x_1+x_2 & 2y_1+y_2\\
\end{bmatrix}
=
\begin{bmatrix}
2x_1+x_2 & 2y_1+y_2\\
x_1 & y_1\\
\end{bmatrix}
= T$
(a) $A = T \cdot M^{-1}$
$A=\begin{pmatrix}2x_{1+x_2}&2y_1+y_2\\ \:\:\:\:x_1&y_1\end{pmatrix}\begin{pmatrix}\frac{y_2}{x_1y_2-y_1x_2}&-\frac{y_1}{x_1y_2-y_1x_2}\\ \:\:\:\:-\frac{x_2}{x_1y_2-y_1x_2}&\frac{x_1}{x_1y_2-y_1x_2}\end{pmatrix}$
$= \begin{pmatrix}\frac{2x_{1+x_2}y_2-x_2\left(2y_1+y_2\right)}{y_2x_1-y_1x_2}&\frac{-2x_{1+x_2}y_1+x_1\left(2y_1+y_2\right)}{y_2x_1-y_1x_2}\\ 1&0\end{pmatrix}$
However the result for $A$ seems wonky. Am I on the right track?
Best Answer
The standard matrix for $T$ is $A=RS$, where $S=\pmatrix{1&2\\0&1}$ and $R=\pmatrix{0&1\\1&0}$.
Can you take it from here?