2001
Write ENM312
312-Q1
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Matrices
January the quadratic forms Q1 = 4x1x2 + 14x1 x3 + 4x2x3 , Q2 = 2x2 + x2 + 2x2 , 2x1 x3 1 2 3 T A X , Q = X T A X , where X is the column vector x x x T . Which in matrix form, Q1 = X 1 2 2 1 2 3 of these is positive de nite? 6 marks T A U is diagonal U is a modal matrix. Verify that Find an orthogonal matrix U such that U 1 U T A2 U is also diagonal and comment. 14 marks Find the eigenvalues, 1 , 2 and 3 of the matrix A, given by 00 1 31 A = @ ,1 0 ,1 A : 3 1 0
312-Q2
3 marks For an arbitrary function ', nd the coe cients p, q and r in ' = p2 + q + r + f Q ; where f is the characteristic function of matrix A. 8 marks Use the result, based on the Cayley-Hamilton theorem, that 'A = pA2 + qA + rI with the same coe cients p, q and r to show that 0 ,1 ,2 ,1 1 01 1 11 0 1 0 ,1 1 1 expAt = 1 et @ 2 4 2 A + e2t @ ,1 ,1 ,1 A + 2 e,3t @ 0 0 0 A : 2 ,1 ,2 ,1 1 1 1 ,1 0 1 9 marks
312-Q3
Write the system of di erential equations 2x01 , x02 + 7x1 , 5x2 = ,19; ,x01 + 2x02 , 5x1 + 7x2 = 17 in the matrix form AX 0 + BX = C ; where A and B are square matrices, X and C are column vectors and ' denotes di erentiation with respect to t. 2 marks Solve the generalized eigenvalue problem AX + BX = 0 i.e. nd the generalized eigenvalues 1 and 2 and the corresponding generalized eigenvectors X1 and X2. 6 marks Verify that X = k1X1e1t + k2X2e2t + B ,1C is a solution of the given system of di erential equations, where k1 and k2 are arbitrary scalar constants. 3 marks Use this to nd the solution of the given di erential equations which satis es the initial conditions x1 0 = 0 and x2 0 = 1. 5 marks Form an orthogonal modal matrix, U , from normalized eigenvectors proportional to X1 and X2 found above. Verify that U T AU and U T BU are both diagonal and comment on the fact that it is possible to nd an orthogonal modal matrix for the matrices A and B . 4 marks
C J S Petrie
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November 2000
Module ENM312 - EXAM January 2001
The given quadratic forms may be written Q1 = xT A1 x and Q2 = xT A2 x, where 00 2 71 0 2 0 ,1 1 A1 = @ 2 0 2 A and A2 = @ 0 1 0 A : 2 marks 7 2 0 ,1 0 2 By inspection, A1 is inde nite. Also almost as easily, since Q2 can be written as a sum of squares, A2 is positive de nite. The characteristic equations are, for A1 , ,3 + 57 + 56 = 0 and, for A2 , 1 , 2 , 4 +3 = 0. The eigenvalues are therefore = ,1; ,7; 8 and = 1; 1; 3 respectively. NOTE
312-Q1 solution
- the eigenvalues are not necessary for the 4 marks here, but will be required later. Marks for lateral thinking for students who make correct inferences without eigenvalues 4 marks The eigenvectors of A1 are 1; ,4; 1T , 1; 0; ,1T and 2; 1; 2T , which are orthogonal. 6 marks Hence it is worth normalizing each and modal matrix so obtained is orthogonal:
0 p1 p 1 B ,18 2 U = B p18 0 B 4 @ 1 ,1
p
18
p
and this will diagonalize A1 . The calculation to show that U T A1 U = diag ,1 ,7 8 is not necessary.
2
2 3 1 3 2 3
1 C C C A
2 marks
0 p1 p 4 p1 1 0 , B p18 18 p18 C @ 0 2 U T A1 U = B 12 0 ,1 C 2 0 @ 2 A 7 2 1 2 2 3 3 3 0 p 1 p4 p 1 1 0 p1 , , B p18 18 p18 C B p18 7 C B ,4 = B ,7 0 B @ 2 2 A @ 18 1 8 16 16
3 3 3
p
It is necessary to verify that A2 is diagonalized:
18
0 1 1 B p118 p2 7 B ,4 2 A B p18 0 0 @ p1 p1 , 18 2 1 1 p 2 0 ,1 2 3C 1C 0 3C=@ 0 0 ,1 2 A
p
2 3
2 3 1 3 2 3
1 C C C A
0
1 0 0 ,7 0 A
8
1 p 2 3 1 3 2 3
Not required
0 p1 p 4 p1 1 0 , B p18 18 p18 C @ 2 U T A2 U = B 12 0 ,1 C 0 @ 2 A ,1 1 2 2 3 3 3 0 p1 p 4 p1 1 0 p1 , B p18 18 p18 C B p18 = B 32 0 ,3 C B ,18 B 4 @ 2 A@ 1 2 1 2
3 3 3
p
The fact that A2 is diagonalized means that it has the same eigenvectors as A1 . In fact, since A2 has a repeated eigenvalue, a set of three eigenvectors is not uniquely determined. 2 marks
312-Q2
18
0 1 B p118 0 ,1 B , 4 1 0 A B p18 0 2 @ p1 1 18 1 p 2 01 2 3C 1C 0 3 C = @0 0 ,1 2 A
p
2 3
, p1
0
2
1 C C C A
4 marks
1 0 0 3 0A
0 1
2
This has the solution
f = ,3 + 7 , 6 = 0 with roots eigenvalue of A 1 = ,3, 2 = 2 and 3 = 1. When = k , f = 0 in the given equation, so that ',3 = 9p , 3q + r '2 = 4p + 2q + r '1 = p + q + r
20p = ,5'1 + 4'2 + ',3 20q = ,5'1 + 8'2 , 3',3 20r = 30'1 , 12'2 + 2',3
The characteristic equation is
solution
3 marks 3 marks
5 marks
C J S Petrie
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November 2000
Module ENM312 - EXAM SOLUTIONS, January 2001
312-Q2
solution continued
A has to satisfy f A = 0 which is the reason that the given result follows from the Cayley-Hamilton theorem. Then we can use the values obtained for p, q and r to write 20'A = 20pA2 + qA + rI = '1 ,5A2 , 5A + 30I + '2 4A2 + 8A , 12I + ',3 A2 , 3A + 2I : 3 marks We calculate 0 8 3 ,1 1 A2 = @ ,3 ,2 ,3 A 1 mark ,1 3 8 and then put this and the value of A in1the previous equation to obtain 0 ,10 ,20 ,10 0 20 20 20 1 0 10 0 ,10 1 20'A = '1 @ 20 40 20 A + '2 @ ,20 ,20 ,20 A + ',3 @ 0 0 0 A : ,10 ,20 ,10 20 20 20 ,10 0 10 5 marks Dividing by 20 and replacing ' by the exponential function gives the result in the question.
and substituting in the matrix equation gives AX 0 + BX = k1e1t 1 AX1 + BX1 + k2e2t 2 AX2 + BX2 + C 1 mark and the fact that the k are eigenvalues and ...