Question

first, consider $(d_{k}circ t_{\langle a,b\
angle})(x,y)$.
$(d_{k}circ t_{\langle a,b\
angle})(x,y)=d_{k}(x + a,y + b)$
$(d_{k}circ t_{\langle a,b\
angle})(x,y)=(kx + ka,ky + kb)$
now, consider $(t_{\langle a,b\
angle}circ d_{k})(x,y)$.
$(t_{\langle a,b\
angle}circ d_{k})(x,y)=t_{\langle a,b\
angle}(kx,ky)$
$(t_{\langle a,b\
angle}circ d_{k})(x,y)=(kx + a,ky + b)$
therefore, a composition of a dilation and a translation be performed in either order and result in the same image.

Explanation:

Step1: Analyze the results of \((D_{k}\circ T_{\langle a,b

angle})(x,y)\) and \((T_{\langle a,b
angle}\circ D_{k})(x,y)\)
We have \((D_{k}\circ T_{\langle a,b
angle})(x,y)=(kx + ka,ky + kb)\) and \((T_{\langle a,b
angle}\circ D_{k})(x,y)=(kx + a,ky + b)\).

Step2: Compare the two results

Since \(ka
eq a\) (unless \(k = 1\)) and \(kb
eq b\) (unless \(k = 1\)), the two compositions \((D_{k}\circ T_{\langle a,b
angle})(x,y)\) and \((T_{\langle a,b
angle}\circ D_{k})(x,y)\) are not equal for \(k
eq1\).

Answer:

cannot