Discrete mathematics i. university question hard. q4) provided in the attachment. cheers! : )

(a) Take any point in the plane . To reflect this point in the -axis, you need to negate the -component, so that

(b) This can be done easily by choosing

as

(c) Computing (very easy since it's diagonal) returns the identity matrix , 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

then

Multiplying by on the left is equivalent to negating the first row of , and on the right to negative the first column of . The third product is equivalent to negating the antidiagonal.