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36. By Exercise 35, theith column ofAisAei, fori= 1,2,3. Therefore,A=147258369.37. We have to solve the systemx1+2x2=2x3=1orx1= 2-2x2x3= 1.Letx2=t. Then the solutions are of the formx1x2x3=2-2tt1, wheretis an arbitraryreal number.38. We will illustrate our reasoning with an example. We generate the “random” 3×3 matrixA=0.1410.5920.6530.5890.7930.2380.4620.6430.383.Since the entries of this matrix are chosen from a large pool of numbers (in our case1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and evenless likely that the whole first column will consist of zeros).This means that we willusually be able to apply Steps 2 and 3 of the Gauss-Jordan algorithm to turn the first41
Chapter 1ISM:Linear Algebracolumn into100; this is indeed possible in our example:0.1410.5920.6530.5890.7930.2380.4620.6430.383-→14.1994.6310-1.680-2.4900-1.297-1.757.Again, it is unlikely that any entries in the second column of the new matrix will be zero.Therefore, we can turn the second column into010.Likewise, we will be able to clear up the third column, so that rref(A) =100010001.We summarize:As we apply Gauss-Jordan elimination to a random matrixA(of any size), it is unlikelythat we will ever encounter a zero on the diagonal. Therefore, rref(A) is likely to have allones along the diagonal.39. We will usually get rref(A) =100a010b001c, wherea, b, andcare arbitrary.40. We will usually have rref(A) =100010001000.(Compare with the summary to Exercise 38.)41. IfAx=bis a “random” system, then rref(A) will usually be100010001, so that we willhave a unique solution.42. IfAx=bis a “random” system of three equations with four unknowns, then rref(A) willusually be100a010b001c(by Exercise 39), so that the system will have infinitely many solutions(x4is a free variable).42
ISM:Linear AlgebraSection 1.343. IfAx=bis a “random” system of equations with three unknowns, then rref[A...b] willusually be100...0010...0001...0000...1, so that the system is inconsistent.44. LetE= rref(A),and note that all the entries in the last row ofEmust be zero, bythe definition of rref.Ifcis any vector inRnwhose last component isn’t zero, thenthe systemEx=cwill be inconsistent.Now consider the elementary row operationsthat transformAintoE,and apply the opposite operations, in reversed order, to theaugmented matrixE...c.You end up with an augmented matrixA...bthatrepresents an inconsistent systemAx=b,as required.45. WriteA= [v1v2. . . vm] andx=x1. . .xm.ThenA(kx) = [v1. . . vm]kx1. . .kxm=kx1v1+· · ·+kxmvmandk(Ax) =k(x1v1+· · ·+xmvm) =kx1v1+· · ·+kxmvm. Thetwo results agree, as claimed.