more efficient way?

is there a more efficient way to translate? I think this is the same method
as Jon's but I needed to write it out to understand it. Anyway, I'm
thinking a mapcar solution. Any ideas?

(defun identity->matrix ( pt mat /
px py pz #0 #1 #2
#00 #01 #02 #03
#10 #11 #12 #13
#20 #21 #22 #23
)
(setq;extract pt
px (car pt)
py (cadr pt)
pz (caddr pt)
)
(setq;extract matrix
#0 (nth 0 mat) #1 (nth 1 mat) #2 (nth 2 mat)
#00 (nth 0 #0) #01 (nth 1 #0) #02 (nth 2 #0) #03 (nth 3 #0)
#10 (nth 0 #1) #11 (nth 1 #1) #12 (nth 2 #1) #13 (nth 3 #1)
#20 (nth 0 #2) #21 (nth 1 #2) #22 (nth 2 #2) #23 (nth 3 #2)
)
(list;rotate/translate pt
(+ (* px #00) (* py #01) (* pz #02) #03)
(+ (* px #10) (* py #11) (* pz #12) #13)
(+ (* px #20) (* py #21) (* pz #22) #23)
)
)

(defun matrix->identity ( pt mat /
px py pz #0 #1 #2
#00 #01 #02 #03
#10 #11 #12 #13
#20 #21 #22 #23
)
(setq;extract pt
px (car pt)
py (cadr pt)
pz (caddr pt)
)
(setq;extract matrix
#0 (nth 0 mat) #1 (nth 1 mat) #2 (nth 2 mat)
#00 (nth 0 #0) #01 (nth 1 #0) #02 (nth 2 #0) #03 (nth 3 #0)
#10 (nth 0 #1) #11 (nth 1 #1) #12 (nth 2 #1) #13 (nth 3 #1)
#20 (nth 0 #2) #21 (nth 1 #2) #22 (nth 2 #2) #23 (nth 3 #2)
)
(setq;translate pt
px (- px #03)
py (- px #13)
pz (- px #23)
)
(list;rotate pt
(+ (* px #00) (* py #10) (* pz #20))
(+ (* px #01) (* py #11) (* pz #21))
(+ (* px #02) (* py #12) (* pz #22))
)
)

 

Oops, mistake in my code...

should be...