Had to find the determinant of a 5x5 matrix using row reduction, and show my steps as part of a homework problem, so I put together some code to do it for me and generate the latex I need while it's at it.

mat.lisp

(defun range (max &key (min 0) (step 1) (runninglist nil))
	(if (<= max min) runninglist (range max :min (+ min step) :step step :runninglist (append runninglist `(,min)))))
(defun entry (mat i j)
	(elt (elt mat i) j))(defun row->latex (row)
	(format t "~a ~{& ~a ~}\\\\~%" (car row) (cdr row))
)
(defun mat->latex (mat)
	(princ "\\begin{vmatrix}")
	(princ #\newline)
	(loop for row in mat do (row->latex row))
	(princ "\\end{vmatrix}")
	(princ #\newline)
)
(defun elim-col (mat col d)
	(let ((eliminated mat) (rows (range (length mat))) (cols (range (length (car mat)))))
		(setf d (* d (entry mat col col)))
		(setf 	(elt eliminated col) 
				(mapcar #'(lambda (x) (/ x (entry mat col col))) (elt eliminated col))
		)
		(unless (= d 1)
			(princ #\=)
			(princ (write-to-string d))
			(princ #\newline)
			(mat->latex eliminated)
		)
		(loop for row in rows do (unless (= row col)
			;(let ((not-print? (= 0 (entry eliminated row col))))
				(setf 	(elt eliminated row) 
						(mapcar #'(lambda (c)	(- 	(entry eliminated row c) 
													(* 	(entry eliminated row col) 
														(entry eliminated col c))))
								cols)
				)
				; Probably more verbose than need be
				;(unless not-print?
				;	(princ #\=)
				;	(unless (= d 1)
				;		(princ (write-to-string d)))
				;	(princ #\newline)
				;	(mat->latex eliminated)
				;)
			;)
		))
		(princ #\=)
		(unless (= d 1) (princ (write-to-string d)))
		(princ #\newline)
		(mat->latex eliminated)
		(list eliminated d)
	)
)
(defun det (mat)
	(mat->latex mat)
	(let ((reduced (list mat 1)) (cols (range (length (car mat)))))
		(loop for col in cols do
			(setf reduced (elim-col (car reduced) col (cadr reduced)))
		)
		(princ #\=)
		(princ (write-to-string (cadr reduced)))
		(cadr reduced)
	)
)