Edwards-Penney Chapter 3, linear algebraic equations and matrices updated 2015 ============================================= problems 3.1-4,6,8,12,16,18,26 ============================================= 3.1-4 5x-6y=1, 6x-5y=10 ===== The answer "x=5, y=4" is the unique solution, as reported by the book. Submitted work is expected to display the solution details, which include documented toolkit steps of swap, combo, mult. Please do not display arithmetic steps. Preferred is a frame sequence which ends with last frame x=5, y=4. Each frame uses exactly one of the 3 rules: combo, swap, mult. 3.1-6 4x-2y=4, 6x-3y=10 ===== The answer is "no solution" because of a signal equation, obtained with the toolkit of 3 rules: combo, swap and mult. A signal equation is a nonzero equation having no variables, like "0=1". Please show the frame sequence details, one operation per display. Point out the signal equation in the last display. 3.1-8 3x-6y=12, 2x-4y=8 ===== Here's a sample of what should be displayed in a solution, using frame sequences and the operations swap, combo, mult. The term "Frame" refers to a video frame, as though the document written was videoed, and only the relevant frames were extracted for view. 3x-6y=12, Frame 1, original problem 2x-4y=8 x-2y=4, Frame 2, mult(row=1,multiplier=1/3) 2x-4y=8 x-2y=4, Frame 3, 0x+0y=0 combo(src=1,target=2,multiplier=-2) x-2y=4, Frame 4, last frame, reduced echelon form 0=0 The last frame algorithm is applied, because there is one lead variable (x) and one free variable (y). Assign invented symbol t1 to the free variable, then back-substitute until all variables are expressed in terms of the invented symbols. Then list the variables, in variable list order, so expressed, as the answer or general solution. x = 4+2y=4+2t1, Final general x = 4+2t1, y = t1 solution is y = t1. 3.1-12 3x+5y- z=13, Solve for the general solution, ====== 2x+7y+ z=28, if it exists, otherwise explain x+7y+2z=32 why there is no solution. The details will be described for equations only, because the matrix interface does not happen until later in the textbook. However, this exercise will likely be solved after a week of exposure to linear algebra and matrices, hence the remarks below about maple. Because maple is used as an assistant, you could first program this system into maple and find the solution. MAPLE ANSWER CHECK. A:=Matrix([[3,5,-1],[2,7,1],[1,7,2]]); b:=<13,28,32>; linalg[linsolve](A,b); C:=; # First Frame linalg[rref](C); # Last Frame # Answer: Au=b has solution u=<1,3,5>. DETAILS. The problem is to write the details by hand, for a frame sequence that starts at the original system and ends at the reduced row echelon form of C, which must be x=1, y=3, z=5. To start, frame 1 is the original system, 3x+5y- z=13, Frame 1 2x+7y+ z=28, x+7y+2z=32 To find frame 2, we identify the appearance of variable x in the system, which is the first variable in the list order x,y,z. If x did not appear, then we would consider y, and then z. If dealing with the trivial system 0=0, 0=0, 0=0, then there are no variables, hence no lead variables, all variables are free variables. We try to make x a lead variable. Rules for display require that lead variable x appear in equation 1. So we arrange this by a swap of equations. x+7y+2z=32, Frame 2, swap(source=1,target=3) 3x+5y- z=13, 2x+7y+ z=28, The plan to make x a lead variable continues. The idea is to remove 3x from equation 2, and 2x from equation 3, using combo. This takes two frames, because we use only one operation per frame. Frame 4 then looks like this: x+7y+2z=32, Frame 4, swap(source=1,target=3) 0+ * * =*, 0+ * * =*, The rest of the work proceeds systematically. Once x is a lead variable, then it appears only in equation 1. If y is to be a lead variable, then it must appear in equation 2 or 3, and we must load equation 2 with the only y appearing in the whole frame. Finally, once x,y are lead variables appearing only once each in the whole frame, which must be equations 1 and 2, by design rules, then we look for the appearance of variable z in the last remaining equation. If absent, then z is a free variable, if present then it can be made into a lead variable. 3.1-16 x-3y+2z=6, x+4y-z=4, 5x+6y+z=20 Find the general solution ====== or else explain why there is no solution. The answer is "no solution" because of a signal equation, obtained with the toolkit of 3 rules: combo, swap and mult. A signal equation is a nonzero equation having no variables, like "0=1". Because there are 3 variables x,y,z, please develop a frame sequence, in which each frame uses just one of the 3 rules, ending with a frame that has a "0=1" equation. MAPLE ANSWER CHECK. The reduced row echelon form of the augmented matrix must show the signal equation. 3.1-18 x+5x+6y=3, 5x+2y-10z=1, 8x+17y+8z=5. Find the general ====== solution or else explain why there is no solution. The book reports "No Solution" which is correct. Display a frame sequence with documented toolkit steps swap, combo, mult, ending in a last frame that contains a signal equation "0=1" followed by the statement "No Solution" or "Inconsistent." MAPLE ANSWER CHECK. The reduced row echelon form of the augmented matrix must show the signal equation. 3.1-26 y''-121y=0, y(0)=44, y'(0)=22, y(x)=A exp(11x)+Bexp(-11x) ====== The constants are A=23, B=21 in the formula y=A exp(11x)+B exp(-11x). Follow 3.1 Example 8. It is not needed to know how to solve the differential equation, no is the differential equation used at all in the solution. This is because the general solution y of the differential equation has been supplied. Section 3.2: 10,14,18,24,26,28 3.2-10 x1-5x2+2x3-7x4+11x5=0, x2-13x3+3x4-7x5=0, x4-5x5=0. Solve ====== the echelon system by back-substitution. BACK SUBSTITUTION. The idea of "back-substitution" is to arrive at the last frame of a frame sequence by aborting the standard algorithm which looks at the variable list from its original ordering, and taking up a new ordering temporarily, namely a reversed variable list, in order to invent a sequence of simplified combos to reach the last frame. The number of combos used is the same as with the original list order. The back-substitution algorithm has limited use for humans and good use on machines, where it could reduce arithmetic overhead. WHAT CAN YOU LEARN FROM ALL THIS? It teaches you to be clever about the choice of the next step in a frame sequence. It is not always best to follow the variable list order, even though that is the plan, eventually. A frame sequence step can take things out of order, to advantage. The signal: you look at the current frame and see that all lead variables are known. A departure from the standard algorithm might then prove to be advantageous. The 3 equations in 5 unknowns have infinitely many solutions. We know this because there are at most three lead variables, hence at least two free variables. The given system is not the last frame in a frame sequence, but very close. All lead variables are already known: x1,x2,x4. Consider the variable list in reverse order: x5,x4,x3,x2,x1. The lead variables are going to be x1,x2,x4. We go down the list x5,x4,x3,x2,x1 starting with x5. It is free, so skip it. Go next to x4. It will be a lead variable, so eliminate x4 from equations 1 and 2. It takes two combos to do it, and these operations are arithmetically identical to "back-substitution". We leave equation 3 in place, as the last equation. Then repeat for x3. It is a free variable, skip it. Go next to x2. It will be a lead variable, so eliminate x2 from equation 1, with a combo. The system is now in reduced row echelon form, that is, the last frame has been found. MAPLE ANSWER CHECK. # We find the last frame directly as the rref of the first frame A:=Matrix([[1,-5,2,-7,11],[1,-13,3,-7],[0,0,0,1,-5]]); linalg[rref](A); # Answer: # Matrix([[1,0,11/8,0,-137/8],[0,1,-1/8,0,11/8],[0,0,0,1,-5]]); x1 + (11/8)x3 - 7x4 - (137/8)x5=0 This is the Last Frame x2 - (1/8)x3 + (11/8)x5=0 Not done. Now apply the x4 - 5x5=0 Last Frame Algorithm The general solution is obtained by applying the Last Frame Algorithm. There is no shortcut for this algorithm, it is not included in back-substitution, nor does the textbook say otherwise. Expected solution steps: Last Frame Algorithm 0. Justify that this is indeed the last frame, by applying the Last Frame Test. 1. Identify lead variables. 2. Identify free variables. Assign invented symbols t1, t2, t3, ... 3. Back-substitute the free variables into the lead variable equations. 4. Report the solution as the list of variables, each followed by an equal sign. Each equal sign is followed by a constant or else an expression in the invented symbols. 3.2-14 3x1-6x2-2x3=1, 2x1-4x2+x3=17, x1-2x2-2x3=-9. Transform to ====== echelon form and solve by back-substitution. Remarks from 3.2-10 apply. Please read them if you skipped that problem. This problem differs from 3.2-10, in that the system is not in triangular form, hence we don't know the lead variables (yet). WHAT CAN YOU LEARN FROM THIS EXAMPLE? Learn that the frame sequence starts with the original system and ends with the last frame, which is uniquely determined, and that back-substitution guides us to find a triangular system in the first frames, in order to organize the final frames into combo steps applied to the lead variables (which were already known from the triangular form). You are actually learning some details of numerical analysis, because in that subject efficiency is important. The 3 equations in 3 unknowns require a number of toolkit steps: combo, swap, mult. Write a frame sequence, one toolkit operation per frame, ending with an intermediate frame in triangular form. This is the important sequence of steps. Then go on to find the last frame, which has one free variable and two lead variables. Justify that this is indeed the last frame, that it passes the Last Frame Test. Then carry out the Last Frame Algorithm to display the solution. MAPLE ANSWER CHECK. # We find the last frame directly as the rref of the first frame A:=Matrix([[3,-6,-2],[2,-4,1],[1,-2,-2]]); b:=<1,17,-9>; C:=; # First Frame linalg[rref](C); # Last Frame # Answer: # Matrix([[1,-2,0,5],[0,0,1,7],[0,0,0,0]]); x1 - 2x2 = 5 This is the Last Frame x3 = 7 Not done. Now apply the 0 = 0 Last Frame Algorithm Expected solution steps: Last Frame Algorithm 0. Justify that this is indeed the last frame, by applying the Last Frame Test. 1. Identify lead variables. 2. Identify free variables. Assign invented symbols t1, t2, t3, ... 3. Back-substitute the free variables into the lead variable equations. 4. Report the solution as the list of variables, each followed by an equal sign. Each equal sign is followed by a constant or else an expression in the invented symbols. # 3.2-14 solved with maple macro(swap=linalg[swaprow]);macro(mult=linalg[mulrow]); macro(combo=linalg[addrow]); A1:=Matrix([[3,-6,-2,1],[2,-4,1,17],[1,-2,-2,-9]]); [3 -6 -2 1] [ ] A1 := [2 -4 1 17] [ ] [1 -2 -2 -9] # last frame in one step linalg[rref](A1); [1 -2 0 5] [ ] [0 0 1 7] [ ] [0 0 0 0] # Frame sequence A1, A2, A3, A4, ... A2:=swap(A1,1,3); [1 -2 -2 -9] [ ] A2 := [2 -4 1 17] [ ] [3 -6 -2 1] > A3:=combo(A2,1,2,-2); [1 -2 -2 -9] [ ] A3 := [0 0 5 35] [ ] [3 -6 -2 1] > A4:=combo(A3,1,3,-3); [1 -2 -2 -9] [ ] A4 := [0 0 5 35] [ ] [0 0 4 28] # Col 1 is the first col of the identity matrix # Proceed to make col 3 the second col of the # identity matrix. Then the last frame is found # which should equal rref(A1) # make a leading 1 in row 2 A5:=combo(A4,3,2,-1); [1 -2 -2 -9] [ ] A5 := [0 0 1 7] [ ] [0 0 4 28] A6:=combo(A5,2,3,-4); [1 -2 -2 -9] [ ] A6 := [0 0 1 7] [ ] [0 0 0 0] A7:=combo(A6,2,1,2); [1 -2 0 5] [ ] A7 := [0 0 1 7] [ ] [0 0 0 0] # last frame test passes linalg[rref](A1); [1 -2 0 5] [ ] [0 0 1 7] [ ] [0 0 0 0] # To finish the solution, translate to equations # and apply the Last Frame Algorithm. # This part of the solution is done by hand. # Answer check A:=Matrix([[3,-6,-2],[2,-4,1],[1,-2,-2]]); b:=<1,17,-9>; linalg[linsolve](A,b); LinearAlgebra[LinearSolve](A,b); 3.2-18 3x1 - 6x2 + x3 + 13x4 = 15 Transform to echelon form, then ===== 3x1 - 6x2 + 3x3 + 21x4 = 21 solve by back-substitution. 2x1 - 4x2 + 5x3 + 26x4 = 23 There are 3 equations and 4 unknowns, therefore at most 3 lead variables and at least one free variable. This implies infinitely many solutions. Like 3.2-14 above, you must display a frame sequence whose intermediate frame is a triangular system. The last frame is a reduced echelon system. Then proceed to apply the Last Frame Algorithm in order to write out the general solution, as in 3.2-14. MAPLE ANSWER CHECK. # We find the last frame directly as the rref of the first frame A:=Matrix([[3, -6, 1, 13],[3, -6, 3, 21],[2, -4, 5, 26]]); b:= <15,21,23>; C:=; # First Frame linalg[rref](C); # Last Frame # Answer: # Matrix([[1, -2, 0, 3, 4], [0, 0, 1, 4, 3], [0, 0, 0, 0, 0]]); x1 - 2x2 + 3x4 = 4 This is the Last Frame x3 + 4x4 = 3 Not done. Now apply the 0 = 0 Last Frame Algorithm 3.2-24 3x+2y=0, 6x+ky=0. Classify according to symbol k the Three ====== Possibilities: (a) Unique solution, (b) No solution, (c) Infinitely many solutions. Because the equations have symbol k appearing in the coefficients, there will in general be infinitely frame sequences, one for each value of k. The last frame of any such sequence can theoretically have a signal equation, or zero free variables, or at least one free variable, which classifies the problem as No Sol, Infinitely Many Solutions or a Unique Solution. The class (a,b,c) the last frame falls into depends on the value of k, and not all classifications need be represented. Display as many complete frame sequences as necessary, each depending on a range of values of k, such that the last frame determines one of the three possibilities. Clearly state assumptions used for each frame sequence. See web references for similar problems, which give further hints on how to organize the work and draw conclusions. 3.2-26 3x+2y=1, 7x+5y=k. Classify according to symbol k the Three ====== Possibilities: (a) Unique solution, (b) No solution, (c) Infinitely many solutions. The advice for 3.2-24 applies, but this problem is considerably simpler, due to there being just one frame sequence reported, valid for all values of k. Conclusion: there are zero free variables and a unique solution. 3.2-28 2x-y+3z=a, x+2y+z=b, 7x+4y+9z=c. Classify according to ====== symbols a,b,c the Three Possibilities: (a) Unique solution, (b) No solutuion, (c) Infinitely many solutions. The basic advice given in 3.2-26 applies. There is just one frame sequence whose last frame has 2 lead variables and one free variable. However, it may have a signal equation if c-2a-3b is nonzero. The unique solution case can never happen, because there always exactly 2 lead variables. MAPLE ANSWER CHECK. # Use maple to find a triangular frame, from which # the result can be deduced. A:=Matrix([[2-1,3],[1,2,1],[7,4,9]]); B:= ; C:=; # First Frame # The maple steps have to be carried out by hand, using # swap, combo, mult [linalg: swaprow, addrow, mulrow] C1:=linalg[swaprow](C,1,2); C2:=linalg[addrow](C1,1,2,-2); C3:=linalg[addrow](C2,1,3,-7); C4:=linalg[addrow](C3,2,3,-2); # Triangular frame is # Matrix([[1,2,1,b],[0,-5,1,-2*b+a],[0,0,0,c-2*a-3*b]]); The conclusion: Two lead variables. Row 3 is zero if and only if the entry c-2*a-3*b is zero. Then the system has no signal equation. 3.3-8 A:=Matrix([[1,-4,-5],[3,-9,3],[1,-2,3]]); Find rref(A). ===== A matrix is in reduced row echelon form <==> the corresponding homogeneous system of equations is a reduced echelon system. There is a simple criterion for testing if a given matrix B is in reduced row echelon form: TEST: Each nonzero row of B has a leading one, and the leading one sits in a column of the identity. Finally, these identity columns appear in natural order as consecutive initial columns of the identity. The textbook reports the answer as the 3x3 identity matrix. Details expected include a frame sequence that starts with matrix A and ends with the identity. Document all steps with detailed information in the form swap(s,t), combo(s,t,c), mult(t,m). 3.3-18 A:=Matrix([[1,-2,-5,-12,1],[2,3,18,11,9],[2,5,26,21,11]]); ===== Find rref(A) The textbook reports the correct answer as a 3x5 matrix B, B=rref(A)=Matrix([[1,0,3,-2,3],[0,1,4,5,1],[0,0,0,0,0]]) Details expected include a frame sequence that starts with matrix A and ends with matrix B. Document all steps with detailed information in the form swap(s,t), combo(s,t,c), mult(t,m). The matrix B is a reduced row echelon form, because the nonzero rows are rows 1 and 2, and each contains a leading one, which sits in a column of the 3x3 identity matrix I. These columns are the first two columns of I in natural order. MAPLE ANSWER CHECK. A:=Matrix([[1,-2,-5,-12,1],[2,3,18,11,9],[2,5,26,21,11]]); linalg[rref](A); 3.3-20 A:=Matrix([[3,6,1,7,13],[5,10,8,18,47],[2,4,5,9,26]]); ====== Find rref(A). The textbook claims the correct answer is a 3x5 matrix B, B=rref(A)=Matrix([[1,2,0,2,3],[0,0,1,1,4],[0,0,0,0,0]]) Details expected include a frame sequence that starts with matrix A and ends with matrix B. Document all steps with detailed information in the form swap(s,t), combo(s,t,c), mult(t,m). The matrix B is a reduced row echelon form, because the nonzero rows are rows 1 and 2, and each contains a leading one, which sits in a column of the 3x3 identity matrix I. These columns are the first two columns of I in natural order. MAPLE ANSWER CHECK. A:=Matrix([[3,6,1,7,13],[5,10,8,18,47],[2,4,5,9,26]]); linalg[rref](A); 3.3-34 ====== The 3x3 matrix forms possible for a reduced echelon form depends on the number of leading ones. Case 1. Three leading ones. Then A:=Matrix([[1,*,*],[0,1,*],[0,0,1]]) initially but to be an rref there has to be zeros above and below a leading one. This forces the form A:=Matrix([[1,0,0],[0,1,0],[0,0,1]]) Case 2. Two leading ones. Then A:=Matrix([[1,*,*],[0,1,*],[0,0,0]]) initially but to be an rref there has to be zeros above and below a leading one. This forces the form A:=Matrix([[1,0,*],[0,1,*],[0,0,0]]) Case 3. One leading one. Then A:=Matrix([[1,*,*],[0,0,0],[0,0,0]]) initially, but there are nochanges to be made, giving A:=Matrix([[1,*,*],[0,0,0],[0,0,0]]) Case 4. No leading one. The A has to be the zero matrix (otherwise, one entry could be made into a leading one). A:=Matrix([[0,0,0],[0,0,0],[0,0,0]]) ============================================= 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. 3.5-14 A:=Matrix([[3,5,6],[2,4,2],[2,3,5]]); ======= Find the inverse of A. Form the matrix C= as frame one of a frame sequence of combo, swap and mult operations ending in last frame equal to rref(C). If the left half of rref(C) is the identity I, then the right half B is the inverse of A. MAPLE ANSWER CHECK. A:=Matrix([[3,5,6],[2,4,2],[2,3,5]]); A^(-1); linalg[rref](>); The problem can be solved entirely in maple by using addrow, swaprow, mulrow to print the frames of the sequence. Then a worksheet print may be submitted. Please include the answer check above, somewhere in your print. 3.5-16 A:=Matrix([[1,-3,-3],[-2,1,2],[2,-3,-3]]); ======= Find the inverse of A. Form the matrix C= as frame one of a frame sequence of combo, swap and mult operations ending in last frame equal to rref(C). If the left half of rref(C) is the identity I, then the right half B is the inverse of A. MAPLE ANSWER CHECK. A:=Matrix([[1,-3,-3],[-2,1,2],[2,-3,-3]]); A^(-1); linalg[rref](>); The problem can be solved entirely in maple by using addrow, swaprow, mulrow to print the frames of the sequence. Then a worksheet print may be submitted. Please include the answer check above, somewhere in your print. 3.5-26 A:=Matrix([[1,5,1],[2,1,-2],[1,7,2]]); ====== B:=Matrix([[2,0,1],[0,3,0],[1,0,2]]); Solve AX=B for X using frame sequences. By hand or maple assist. Form the matrix C:=; as frame one of a frame sequence of combo, swap and mult operations ending in last frame equal to rref(C). If the left half of rref(C) is the identity I, then the right half D is the inverse of A multiplied against B, that is, D = inverse(A) B. MAPLE ANSWER CHECK. with(linalg): A:=Matrix([[1,5,1],[2,1,-2],[1,7,2]]); B:=Matrix([[2,0,1],[0,3,0],[1,0,2]]); A^(-1).B; linalg[rref](); The problem can be solved entirely in maple by using addrow, swaprow, mulrow to print the frames of the sequence. Then a worksheet print may be submitted. Please include the answer check above, somewhere in your print. 3.5-34 Show that a diagonal matrix is invertible if and only if each ====== diagonal element is nonzero. The result for dim(A)=3 is A=diag(a,b,c) and A invertible if and only if abc is not zero. The result for dim(A)=n is A=diag(a1,a2,...,an) and A invertible if and only if a1*a2*...*an is not zero. You must present a proof for dim(A)=n. A dim(A)=3 proof earns 40% or less. Resource: Theorem page 190, which says A is invertible if and only if rref(A)=I. For dim(A)=3, the argument goes like this: I. If abc=0, then A has a row of zeros, hence rref(A) has a row of zeros and then rref(A) is not the identity I. By page 190, A fails to be invertible. II. If abc is not zero then three multiply rules can be applied to reduce matrix A to I, hence rref(A)=I. Page 190 implies A is invertible. III. In case II, the inverse of A can be found by using the same three mult rules on the augmented matrix C= to give rref(C)= where B=diag(1/a,1/b,1/c). REMARKS on 3.5-34, Spring 2012 ====== Other proofs are possible using just the definition of invertible matrix. The problem can be distilled to proving these two statements: (a) Given A=diag(a1,a2,...,an) and B=diag(1/a1,1/a2,...,1/an) and a1,...,an are not zero, then AB=I and therefore B is the inverse of A. (b) Given A=diag(a1,a2,...,an) and B is inverse to A, that is , AB=BA=I, then a1,...,an are not zero. Statement (a) is proved by direct verification. This means that A and B are written with "..." notation next to each other and multiplication is carried out to get the diagonal matrix I. Statement (b) is proved by writing out the equation AB=I using a general matrix B with entries b_{ij}. The equations a_{11}b_{11}=1,...,a_{nn}b_nn}=1 are obtained. These equations imply that a_{11} is not zero, ..., a_{nn} is not zero. This problem may be submitted in place of 44. 3.5-44 Show that every invertible matrix is the product of elementary ====== matrices. The proof requires the use of (15), page 194, plus the fact that an inverse of an elementary matrix is also an elementary matrix. Expected details repeat parts of the proof of Theorem 6 [page 194], obtaining (15) for the present setting. Update 2012: Extra credit problem Xc3.5-44a replaces 3.5-44. The logic here is that problem 44 is a proof of three pages, whereas the Xc problem is a (a) a little bit of theory and (b) a computation. 3.6-6 Expand a 5x5 determinant by cofactors. ===== The challenge of cofactor expansion of a 5x5 determinant is to begin each expansion opportunity using a row or column that contains as many zeros as possible. The best one for the 5x5 is row 3, which contains 4 zeros. The expansion then gives (+1)(5)(a 4x4 determinant). The checkerboard sign (+1) is found by marking from th eupper left corner of the 5x5 determinant, saying "plus - minus - plus - minus -plus" in going from the (1,1)-position to the (3,3)-poition [where 5 is located]. The progress marked, to obtain a 4x4 determinant to work on, then the next step chooses a cofactor expansion of the 4x4 on the last row, which contains 3 zeros. This time the answer is (-1)(2)(3x3 determinant). The process continues, with a row one expansion of the 3x3 [it has 2 zeros] to get a 2x2 determinant, which is evaluated using the 2x2 Sarrus' rule. MAPLE ANSWER CHECK. A:=Matrix([[3,0,11,-5,0],[-2,4,13,6,5], [0,0,5,0,0],[7,6,-9,17,7],[0,0,8,2,0]]); linalg[det](A); # answer = 60 3.6-20 A=4x4 matrix, find |A| using the four rules, triangular, combo, ====== swap, mult. The method of elimination is a synonym for the Four Rules (1) The triangular rule, or 1-arrow Sarrus' rule; (2) Swap [swap(s,t) changes the sign of the determinant]; (3) Mult [mult(t,m) divides the determinant by m]; (4) Combo [combo(s,t,c) leaves the determinant unchanged]. The last 3 rules are equivalent to one rule, a special case of the determinant product theorem, det(A) = det(EA)/det(E). The objective is to apply combo to the matrix to get a triangular form, then apply the one-arrow Sarrus' rule to evaluate the determinant. 3.6-30 Solve a 3x3 system by Cramer's Rule. ===== Cramer's rule for this problem is x1 = det(A1)/det(A), x2 = det(A2)/det(A), x3 = det(A3)/det(A). The symbols are: A = matrix whose rows are [1,4,2], [4,2,1], [2,-2,-5]. b = column vector with entries 3, 1, -3 A1 = matrix A with col(A,1)=b A2 = matrix A with col(A,2)=b A3 = matrix A with col(A,3)=b You are expected to evaluate the four determinants det(A), det(A1), det(A2), det(A3), simplify the fractions and report the answer x1, x2, x3. MAPLE ANSWER CHECK. A:=Matrix([[1,4,2],[4,2,1],[2,-2,-5]]); b:=<3,1,-3>; linalg[linsolve](A,b); # Answer: x1 -1/7, x2 = 9/14, x3 = 2/7 3.6-32 Solve a 3x3 system by Cramer's Rule. ===== Cramer's rule for this problem is x1 = det(A1)/det(A), x2 = det(A2)/det(A), x3 = det(A3)/det(A). The symbols are: A = matrix whose rows are [3,4,-3], [3,-2,4], [3,2,-1]. b = column vector with entries 5,7,3 A1 = matrix A with col(A,1)=b A2 = matrix A with col(A,2)=b A3 = matrix A with col(A,3)=b You are expected to evaluate the four determinants det(A), det(A1), det(A2), det(A3), simplify the fractions and report the answer x1, x2, x3. MAPLE ANSWER CHECK. A:=Matrix([[3,4,-3], [3,-2,4], [3,2,-1]]); b:=<5,7,3>; linalg[linsolve](A,b); # Answer: x1 = -7/3, x2 = 9, x3 = 8 3.6-40 Apply the adjugate formula for the inverse to find the inverse of ====== the given matrix A. A:=Matrix([[2,4,-3], [2,-3,-1], [-5,0,-3]]); This problem will help you to understand the adjugate formulas inverse(A) = adj(A)/det(A), adj(A) = adjugate or adjoint of A = transpose of the matrix of cofactors of A The following maple code can be used as an answer check, but not as a solution to the problem. The problem solution is expected to be hand-written, with all 2x2 determinant details documented. The expected presentation is a formula for adjugate(A) as the transpose of a big matrix whose 9 entries are written as 2x2 determinants multiplied by checkerboard signs [cofactor = sign x minor]. For presentation details, follow Example 11 Section 3.6 in E&P. MAPLE ANSWER CHECK. A:=Matrix([[2,4,-3], [2,-3,-1], [-5,0,-3]]); linalg[adjoint](A); linalg[det](A); linalg[inverse](A); 3.6-60 Let B[1]:=2 (a number), B[2]:=Matrix([[2,1],[1,2]]); ===== B[3]:=Matrix([[2,1,0],[1,2,1],[0,1,2]]); B[4]:=Matrix([[2,1,0,0],[1,2,1,0],[0,1,2,1],[0,0,1,2]]); B[5]:=Matrix([[2,1,0,0,0],[1,2,1,0,0], [0,1,2,1,0],[0,0,1,2,1],[0,0,0,1,2]]); and so on. Part (a). Expand along the first row to show that B[n]=2B[n-1]-B[n-2] for n>=3. Intuition to solve the problem can be obtained by writing out explicitly the determinants B1, B2, B3, B4, B5. Subject B5 to cofactor expansion along row 1 to get the recursion identity B5=2*B4-B3 of 3.6-60a. The work submitted must be for dimension n, for full credit. Examples are not a proof of part (a). Use "..." notation to write out B_{n}, B_{n-1} and B_{n-2}. Cite steps where cofactor expansion by row and by column are used. Part (b). Prove by induction that B[n]=n+1 for n>=2. Induction is explained in the appendix to the Varberg calculus text which was used for Calc I,II,III. Part 3.6-60b is expected to be solved using the theory of mathematical induction. Graders will award two possible grades: 50, 100. For 100 percent, the work must use induction methods. For 50 percent, some drastic unforgivable error has to occur, for example, submitting a work which does examples but no induction, or choosing a sequence of details that bears little or no relationship to the expected details. Exposition details (a) ========= Write out determinant B_n on paper using "..." notation. At least six (6) columns should be written, to give enough detail. Then use cofactor expansion along row 1 to produce 2 determinants of order n-1. Finally, use column 1 cofactor expansion on the second determinant to produce a determinant of order n-2. All of this can fit one one sheet of paper. Proof details (b) ========= While many possible definitions exist for statement P_n in math induction, all of them have to have two or three equations in order apply the recursion formula of (a): (a) B_k = 2 B_{k-1} - B_{k-2} for k > 3 The problem is that to use (a), two equations are needed for expansion of the RHS. Bear in mind that P_n = { B_n = n+1} does not work for two reasons: first, P_1 = { B_1=2} and B_1 is not defined, only B_2 and higher; second, not enough equations appear in P_n to expand the RHS of (a). Something like P_n = { B_{n} = n+1 and B_{n-1} = n and B_{n-2} = n-1 } does work. The details do not appear here, but are expected on submitted work. Here, n >= 3, so the first statement is P_3, where P_3 = { B_{3} = 4 and B_{2} = 3 and B_{1} = 2 } Show the first statement P_3 is true. Then show that if one statement P_n is true then the next statement P_{n+1} is true, where P_n = { B_{n} = n+1 and B_{n-1} = n and B_{n-2} = n-1 } [assumed true] P_{n+1} = { B_{n+1} = n+2 and B_{n} = n+1 and B_{n-1} = n } [to be proved true] ===end===