Problem 3.5-44 is a proof. It will not be due, but the following substitution is due. ======================================================================= Problem 3.5-44 [replacement] Invent a paricular 3x3 invertible matrix A1 and display a frame sequence A1 to A6 (or slightly longer) involving documented steps of combo, swap and mult (one of each at least). Then write the frame sequence in the form A6 = E5 E4 E3 E2 E1 A1 where E1, ..., E5 are the elementary matrices representing the combo, swap and mult operations. Finally, check your answer by multiplying out the RHS of the above identity, showing the multiplication gives A6 (which should be rref(A1)=I). ======================================================================= Example. The same problem but for 2x2 A1. [0 3] A1 = [2 4] [0 3] [1 0 ] A2 = [1 2] mult(2,1/2), E1 = [0 1/2] [1 2] [0 1] A3 = [0 3] swap(1,2), E2 = [1 0] [1 2] [1 0 ] A4 = [0 1] mult(2,1/3), E3 = [0 1/3] [1 -2] [1 0] combo(2,1,-2), E4 = [0 1] A5 = [0 1] Then A5 = E4 A4 = E4 E3 A3 = E4 E3 E2 A2 = E4 E3 E2 E1 A1 Then multiplying out the four elementary matrices by hand gives [-2/3 1/2] E4 E3 E2 E1 = [ 1/3 0 ] and then [1 0] E4 E3 E2 E1 A1 = [0 1] = I This last check can be done in maple by defining each 2x2 elementary matrix, e.g., A1:=matrix([[0,3],[2,4]]); and then with(linalg): evalm(E4&*E3&*E2&*E1&*A1); The last line gives the identity, which is A5, and that completes the answer check. =====================================================================