# Recitation 4A

Problem 1: Let \mathbf{b}=\begin{bmatrix}-1 \\ 0 \\ 0\end{bmatrix} and let A be the matrix \begin{bmatrix}1 & -3 & 5 & -5\\0 & 1 & -3 & 5\\2 & -4 & 4 & -4\end{bmatrix}. Is \mathbf{b} in the range of the linear transformation \mathbf{x}\mapsto A\mathbf{x}? Why or why not?

Solution: Because \mathbf{b} is in the range of the linear transformation \mathbf{x}\mapsto A\mathbf{x}‘ is equivalent to say \mathbf{b} is a linear combination of the columns of A‘. Therefore it is enough to check if the augmented matrix \begin{bmatrix}1 & -3 & 5 & -5 & -1\\0 & 1 & -3 & 5 & 0\\2 & -4 & 4 & -4 & 0\end{bmatrix} is consistent. By row reductions, its echelon form is
\begin{bmatrix}1 & -3 & 5 & -5 & -1\\0 & 1 & -3 & 5 & 0\\0 & 0 & 0 & -4 & 2\end{bmatrix} which is consistent. Therefore \mathbf{b} is in the range of the linear transformation.

Problem 2: Let T: \mathbb{R}^2\to\mathbb{R}^2 be the transformation that first performs a horizontal shear that maps e_2 into e_2-e_1/2 and then reflects the result through the x_2-axis. Assuming T is linear, find its standard matrix.

Solution: It is enough to find Te_1 and Te_2. Noticing that T transforms e_1 to e_1 via the horizontal shear first and then transforms e_1 to -e_1 via the reflection, Te_1 = -e_1=\begin{bmatrix}-1\\ 0\end{bmatrix}. On the other hand, T transforms e_2 to e_2 - e_1 / 2 via the horizontal shear first and then transforms e_2-e_1/2 to e_2+e_1/2 via the reflection. Therefore Te_2=e_2+e_1/2=\begin{bmatrix}1/2\\1\end{bmatrix}. Hence the standard matrix of T is \begin{bmatrix}-1 & 1/2\\0 & 1\end{bmatrix}.