(a) Take any point [tex](x,y)[/tex] in the plane [tex]\mathbb R^2[/tex]. To reflect this point in the [tex]y[/tex]-axis, you need to negate the [tex]x[/tex]-component, so that
[tex]\mathbf T(\vec x)=\begin{bmatrix}-x\\y\end{bmatrix}[/tex]
(b) This can be done easily by choosing
[tex]\mathbf T=\begin{bmatrix}-1&0\\0&1\end{bmatrix}[/tex]
as
[tex]\begin{bmatrix}-1&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-x\\y\end{bmatrix}[/tex]
(c) Computing [tex]\mathbf T^2[/tex] (very easy since it's diagonal) returns the identity matrix [tex]\mathbf I_2[/tex], which as a linear transformation returns whatever it is applied to. Geometrically, we're applying the same reflection twice, which returns the original vector to its starting value/position in the plane.
(d) If
[tex]\mathbf A=\begin{bmatrix}a&b\\c&d\end{bmatrix}[/tex]
then
[tex]\mathbf{TA}=\begin{bmatrix}-a&-b\\c&d\end{bmatrix}[/tex]
[tex]\mathbf{AT}=\begin{bmatrix}-a&b\\-c&d\end{bmatrix}[/tex]
[tex]\mathbf{TAT}=\begin{bmatrix}a&-b\\-c&d\end{bmatrix}[/tex]
Multiplying [tex]\mathbf A[/tex] by [tex]\mathbf T[/tex] on the left is equivalent to negating the first row of [tex]\mathbf A[/tex], and on the right to negative the first column of [tex]\mathbf A[/tex]. The third product is equivalent to negating the antidiagonal.
