We have seen that any matrixMcan be put into reduced row echelonform via a sequence of row operations, and we have seen that any row op-eration can be emulated with left matrix multiplication by an elementarymatrix.Suppose that RREF(M) is the reduced row echelon form ofM.Then RREF(M) =E1E2. . . EkMwhere eachEiis an elementary matrix.ExampleThe example from Lecture 3 on Elementary Row Operations,written using elementary matrices, would look like:100301010013=10001-2001101201000110001000131-60010001·1000-10001100-110001300010001001010100003915-2213203Or, using shorthand, we have:RREF(M) =S23(-2)S13(12)R3(13)S12(-6)R2(-1)S21(-1)R1(3)E13MOf course, this example shows that you can use elementary matrices to re-duce augmented or non-square matrices (when speaking of the determinant,however, we are only concerned with square matrices).What is the determinant of a square matrix in reduced row echelon form?•IfMis not invertible, then some row of RREF(M) contains only zeros.Then we can multiply the zero row by any constantλwithout chang-ingM; by our previous observation, this scales the determinant ofMbyλ. Thus, ifMis not invertible, det RREF(M) =λdet RREF(M),and so det RREF(M) = 0.•Otherwise, every row of RREF(M) has a pivot on the diagonal; sinceMis square, this means that RREF(M) is the identity matrix. ThenifMis invertible, det RREF(M) = 1.•Additionally, notice that det RREF(M) = det(E1E2. . . EkM).Thenby the theorem above, det RREF(M) = det(E1). . .det(Ek) detM. SinceeachEihas non-zero determinant, then det RREF(M) = 0 if and onlyif detM= 0.