Math 43, Fall 2007B. DodsonWeek 9:Monday: Finish Suggested Hw8, start on Hw9 material1.Determinants2.Properties of Dets3. Eigenvalues and Eigenvectors (2-by-2 in 4.1;then n-by-n in 4.3)——–We compute det215423951using the(first) row expansion (by minors):det(A) = 2flflflfl2351flflflfl-1flflflfl4391flflflfl+ 5flflflfl4295flflflfl= 2(2-15)-(4-27) + 5(20-18)= 2(-13)-(-23) + 5(2) =-26 + 33 = 7.———–ProblemReduceA=2135301241435253to an upper triagularmatrix and use the reduction to finddet(A).Solution:A→-1123301241435253(r1→r1-r2)→-112303711051215071518(r2→r2+ 3r1, r3→r3+ 4r1, r4→r4+ 5r1)
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.
33. Eigenvalues and Eigenvectors—————A vectorv= 0 inRnis aneigenvector with eigenvalueλofann-by-nmatrixAifAv=λv.We re-write the vector equation as (A-λIn)v= 0,which is a homogeneous system with coefmatrix (A-λI),and we wantλso that the system has a non-trivial solution.