============================================= S2013 notes on problems 3.4-20,22,30,34,36,40 ============================================= 3.4-20 x1 - 3 x2 + 0 x3 + 0 x4 + 7 x5 = 0, x3 + 0 x4 - 2 x5 = 0, x4 - 10 x5 = 0. A:=Matrix([[1,-3,0,0,7],[0,0,1,0,-2],[0,0,0,1,-10]]); Solve Au=0 in vector form. The system of equations is already a reduced echelon system, that is, the augmented matrix from the first frame equals the last frame, which is the RREF. This is a problem with three lead variables and two free variables. It is the infinitely many solution case. The algorithm used is as follows: Assign symbols t1, t2 to the free variables [book: symbols s,t] Back-substitute the invented symbols into the equations of the last frame. Then solve for the lead variables. Report the variable list equalities, in variable list order, as the general scalar soluton. Write the scalar solution as a vector solution of the form u = t1 v1 + t2 v2, where u, v1, v2 are column vectors. Argue that v1 = partial of u w/r/t t1 and v2 = partial of u w/r/t t2. MAPLE ANSWER CHECK. A:=Mmatrix([[1,-3,0,0,7],[0,0,1,0,-5],[0,0,0,1,-10]]); linalg[linsolve](A,vector([0,0,0])); 3.4-22 System x1 - x2 + 0 x3 + 7 x4 + 3 x5 = 0, x3 - x4 - 2 x5 = 0. A:=Matrix([[1,-1,0,7,3],[0,0,1,-1,-2]]); Find the solution of Au=0 in vector form. The system of equations is already a reduced echelon system, that is, in augmented matrix from the first frame equals the last frame, which is the RREF. This is a problem with three lead variables and two free variables. It is the infinitely many solution case. The algorithm used is as follows: Assign symbols t1, t2 to the free variables [book: symbols s,t] bBack-substitute the invented symbols into the equations of the last frame. Then solve for the lead variables. bReport the variable list equalities, in variable list order, as the general scalar soluton. Write the scalar solution as a vector solution of the form u = t1 v1 + t2 v2, where u, v1, v2 are column vectors. Argue that v1 = partial of u w/r/t t1 and v2 = partial of u w/r/t t2. MAPLE ANSWER CHECK. A:=matrix([[1,-3,0,0,7],[0,0,1,0,-5],[0,0,0,1,-10]]); linsolve(A,vector([0,0,0])); 3.4-29 for A=Matrix([[a,b],[c,d]]) show that A^2-(trace(A))A + det(A)I=0 where trace(A)=a+d and det(A)=ad-bc This result is called the Cayley-Hamilton theorem. It is the basis for solving system of differential equations of the form x'=Ax, where A is a square matrix. The general theorem reads "A square matrix A satifies its own characteristic equation." The result is given in 3.4-29 for the special case of a 2x2 matrix, however Cayley-Hamilton holds for nxn matrices. Special Case 2x2 matrix A (-r)^2 + trace(A)(-r) + det(A) = 0 Characteristic equation (-A)^2 + trace(A)(-A) + det(A)I = 0 Cayley-Hamilton identity The characteristic equation is the formal expansion det(A-rI)=0. In college algebra, 2x2 and 3x3 determinants were studied. The topic of determinant theory appears in 3.6 of the textbook. The proof is done bare-handed by expanding the two sides of the proposed identity, that is, compute A^2 by hand, then compute by hand (a+d)A+(ad-bc)I, showing in the end the two 2x2 matrices so computed are equal. 3.4-30 Let A:=Matrix([[2,1],[1,2]]); Compute using only scalar multiply and matrix add the products A^2, A^3, A^4, A^5, using the Cayley-Hamilton identity of 3.4-29. See 3.4-29 above for the Cayley-Hamilton formula for a general 2x2 matrix A. For the given matrix A, this formula is r^2 -4r + 3 = 0 Characteristic equation A^2 -4A + 3I = O Cayley-Hamilton identity Then A^2 = 4A - 3I, A^3 = A(A^2) = A(4A - 3I) = 4A^2-3AI = 4A^2 - 3A because AI=A = 4(4A-3I)-3A = 13A - 12I Because of these formulas, the powers A^2, A^3, A^4, A^5 can be computed from linear combinations of A and I. No matrix multiply is needed. MAPLE ANSWER CHECK. A:=Matrix([[2,1],[1,2]]); A^5; Then maple will print A^5, which if it matches your A^5 done the other way, proves that your computation is error free. 3.4-34 Find a 2x2 matrix A:=Matrix([[a,b],[c,d]]) with |a|=|b|=|c|=|d|=1 such that A^2=Matrix([[0,0],[0,0]]). Use 3.4-29, the Cayley Hamilton theorem as follows. The equation A^2 = O is the same as A^2 -(a+d)A+(ad-bc)I=O where the components a,b,c,d of A have been chosen so that a+d=0 and ad-bc=0. This means the trace of A is zero and the determinant of A is zero. There are lots of matrices A with these conditions satisfied: pick one and show how you got it. ANSWER CHECK. This is a proof, so there is formally no answer check, but you should display steps that verify the equation A^2=0. 3.4-36 Find a 2x2 matrix A:=Matrix([[a,b],[c,d]]) with a=d=0 such that A^2=Matrix([[1,0],[0,1]]). Use 3.4-29, the Cayley Hamilton theorem as follows. The equation A^2 = I because of A^2 -(a+d)A+(ad-bc)I=O and a=d=0 implies I-bcI=0. This happens if and only 1-bc=0. There are lots of matrices A:=Matrix([[0,b],[c,0]]) with this condition satisfied: pick one and show how you got it. ANSWER CHECK. This is a proof, so there is formally no answer check, but you should display steps that verify the equation A^2=I. 3.4-40 We re-state the problem in slightly different language. (a) Given A x_0 = 0 and A x_1 = b, then define x = x_0 + x_1. We show that A x = b, as follows: (LHS of eq Ax=b) = Ax = A(x_0 + x_1) substitute = Ax_0 + Ax_1 matrix multiply rules = 0 + b use given equations = b vector toolkit property = (RHS of eq Ax=b) The proof is complete. (b) Give A x_1 = b and A x_2 = b, then define x = x_1 - x_2. Show that A x = 0. The details, left to you, parallel the matrix algebra details of (a). In particular, do not break A and x into components; use only formal matrix algebra rules, as in (a) above. ==========