2→-112303711051215011-4(r4→r4-2r2)——–→-1123011-405121503711(r2↔r4)→-1123011-40073500423(r3→r3-5r2, r4→r4-3r2)→-1123011-400-1-1100423(r3→r3-2r4)→-1123011-400-1-11000-21=U(r4→r4+ 4r3).Now we have detU= (-1)(1)(-1)(-21) =-21,and detA= (-1) detU= 21,since all of the 3rdelementary operations change the determinant by afactor of (1) – so no change - there are no 2ndEROs and exactly one 1st ERO, with each 1st EROchanging the determinant by a factor of (-1).Notice that, for the problem of finding the value ofthe determinant, we do not need a reduced row echelonor 1’s on the diagonal, and stop at a triangular form.