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Math 246, Sample Problem Solutions for Third In-Class Exam
(1) Consider the matrices
A =
3 1 − i 2 + i 4
, B =
Compute the matrices (a) AT^ , Solution: The transpose of A is given by
AT^ =
3 2 + i 1 − i 4
(b) A, Solution: The conjugate of A is given by
A =
3 1 + i 2 − i 4
(c) A∗, Solution: The adjoint of A is given by
A∗^ =
3 2 − i 1 + i 4
(d) 2A − B, Solution: The difference of 2A and B is given by
2 A − B =
6 2 − i 2 4 + i 2 8
0 − 5 − i 2 −1 + i 2 2
(e) AB, Solution: The product of A and B is given by
AB =
3 1 − i 2 + i 4
3 · 6 + (1 − i) · 5 3 · 7 + (1 − i) · 6 (2 + i) · 6 + 4 · 5 (2 + i) · 7 + 4 · 6
23 − i 5 27 − i 6 32 + i 6 38 + i 7
(f) B−^1. Solution: Observe that it is clear that B has an inverse because
det(B) = det
The inverse of B is given by
B−^1 =
This may be computed in a number of ways. Here are three. 1
First, it can be computed by applying elementary row operations to transform the augmented matrix (B | I) into (I | B−^1 ) as follows: ( B
∣ I)^ =
I
∣ B−^1 )^.
Second, because B is a two-by-two matrix, its inverse can be computed directly from the formula ( a b c d
ad − bc
d −b −c a
whenever ad − bc 6 = 0.
This formula is just formula (24) on page 352 of the book specialized to the two-by-two case. When applied to B it yields
B−^1 =
Finally, the inverse of B may be computed directly from its definition by seeking a, b, c, and d such that ( 1 0 0 1
a b c d
6 a + 7c 6 b + 7d 5 a + 6c 5 b + 7d
Then a and c are found by solving the two-by-two system 6 a + 7c = 1 , 5 a + 6c = 0 , which gives a = 6 and c = −5. Similarly b and d are found by solving the two-by-two system 6 b + 7d = 0 , 5 b + 6d = 1 , which gives b = −7 and d = 6. You thereby find that
B−^1 =
a b c d
(2) Consider the matrix
A =
(a) Find all the eigenvalues of A. Solution: The eigenvalues are the roots of the equation
0 = det(A − λI) = det
5 − λ − 3 − 3 5 − λ
= (5 − λ)^2 − 32.
The eigenvalues are therefore λ = 5 ± 3, or simply 2 and 8.
Solution: First, a remark. By any method you can compute
det
This does not mean however that there is no solution. All it means is that there might be no solution. Whether there is a solution or not depends on the right-hand side of the system. For example, if the numbers on the right-hand side are 1, 2, and 1 rather than 1, 1, and 1, then the system would have the solution x 1 = 1, x 2 = 0 and x 3 = 0. In fact, the given system has no solution. There are many ways to see this. One of the simplest is to add the first and third equations to obtain 2 x 1 + x 2 + x 3 = 2 , which clearly contradicts the second equation. Done. You can find similar contradictions by taking other combinations of the equations. For example, subtracting the first from the second contradicts the third. Done. Alternatively, you can try to solve the system by applying elementary row operations to the augmented matrix. This gives
The last two rows on the right-hand side are clearly contradictory, whereby the system has no solution. Done. If you did not see the contradiction above and continued to apply ele- mentary row operations, you would get
3 0 − 3 3
3 0 1 − 1
3 0 0 0
At this point the last row clearly shows that the system has no solution. You have arrived at the right answer, but it took you too long to get there!
(4) Solve each of the following initial-value problems. (a) d dt
x 1 x 2
x 1 x 2
x 1 (0) x 2 (0)
Solution: First, you must compute the characteristic polynomial p(z) of the coefficient matrix
A =
Because A is 2×2 this can either be done as p(z) = z^2 − tr(A) z + det(A) = z^2 − (1 − 2)z + (− 2 − 4) = z^2 + z − 6 ,
or as
p(z) = det(A − zI)
= det
1 − z 1 4 − 2 − z
= (1 − z)(− 2 − z) − 4
= z^2 − (1 − 2)z − 2 − 4 = z^2 + z − 6.
By either method, after factoring you obtain
p(z) = (z + 3)(z − 2) ,
whereby the eigenvalues of A are −3 and 2.
Now there are several approaches you can take. The approach used in class goes as follows. First, compute the exponential matrix exp(At). Because A has two simple real roots one has that
exp(At) =
[
(2I − A) e−^3 t^ + (A + 3I)e^2 t
]
[(
e−^3 t^ +
e^2 t
]
The solution of the initial-value problem is then given by
x(t) = exp(At)
[(
e−^3 t^ +
e^2 t
]
[(
e−^3 t^ +
e^2 t
]
The approach used in the book goes as follows. First compute eigen- vectors associated with the eigenvalues −3 and 2 respectively: ( 1 − 4
, and
The general solution is thereby found to be
x(t) = c 1
e−^3 t^ + c 2
e^2 t^.
The initial condition then leads to the equations
c 1 + c 2 = 2 , − 4 c 1 + c 2 = − 1.
These are then solved to find c 1 = 3/5 and c 2 = 7/5. Hence,
x(t) =
e−^3 t^ +
e^2 t^.
Notice that the details of finding the eigenvectors and of solving for c 1 and c 2 are omitted above. When these details are added it should be clear to you that this approach takes much longer than the approach used in class.
These are then solved to find c 1 = − 1 /4 and c 2 = 3/4. Hence,
x(t) =
e−^3 t^ +
e^2 t^.
Notice that the details of finding the eigenvectors and of solving for c 1 and c 2 are omitted above. When these details are added it should be clear to you that this approach takes much longer than the approach used in class.
(5) Find a greneral solution for each of the following systems. (a) d dt
x 1 x 2
x 1 x 2
Solution: The characteristic polynomial of the coefficient matrix A is
p(z) = z^2 − 4 z + 8 = (z − 2)^2 + 4.
This has the roots z = 2 ± i2. Now there are several approaches you can take. The approach used in class goes as follows. First, compute the exponential matrix exp(At). Because A has the simple conjugate pair of roots 2 ± i2 one has that
exp(At) = Ie^2 t^ cos(2t) + (A − 2 I)e^2 t^
sin(2t) 2
=
e^2 t^ cos(2t) +
e^2 t^
sin(2t) 2
= e^2 t
cos(2t) − sin(2t) −2 sin(2t) sin(2t) cos(2t) + sin(2t)
A general solution is therefore
x(t) =
x 1 (t) x 2 (t)
= exp(At)
c 1 c 2
= e^2 t
c 1 [cos(2t) − sin(2t)] − 2 c 2 sin(2t) c 1 sin(2t) + c 2 [cos(2t) + sin(2t)]
(b) d dt
x y
x y
Solution: The characteristic polynomial of the coefficient matrix A is
p(z) = z^2 − 3 z − 4 = (z + 1)(z − 4).
This has the roots z = −1 and z = 4. Now there are several approaches you can take. The approach used in class goes as follows. First, compute the exponential matrix exp(At).
Because A has the two simple real roots −1 and 4 one has that
exp(At) =
[
(4I − A) e−t^ + (A + I)e^4 t
]
[(
e−t^ +
e^4 t
]
6 e^4 t^ − e−t^3 e^4 t^ − 3 e−t 2 e−t^ − 2 e^4 t^6 e−t^ − e^4 t
A general solution is therefore given by
x(t) =
x 1 (t) x 2 (t)
= exp(At)
c 1 c 2
6 e^4 t^ − e−t^3 e^4 t^ − 3 e−t 2 e−t^ − 2 e^4 t^6 e−t^ − e^4 t
c 1 c 2
c 1 [6e^4 t^ − e−t] + c 2 [3e^4 t^ − 3 e−t] c 1 [2e−t^ − 2 e^4 t] + c 2 [6e−t^ − e^4 t]
(6) Sketch phase-plane portraits for each of the following systems. State the type and stability of the origin. (a) d dt
x 1 x 2
x 1 x 2
Solution: The characteristic polynomial of the coefficient matrix is p(z) = z^2 + z + (− 2 − 4) = z^2 + z − 6 = (z + 3)(z − 2). This has the roots z = −3 and z = 2 with corresponding eigenvectors given by (^) ( 1 − 4
and
The origin thereby is a saddle and is unstable. The portrait is attract- ing along the line y = − 4 x and repelling along the line y = x. (b) d dt
x 1 x 2
x 1 x 2
Solution: The characteristic polynomial of the coefficient matrix is p(z) = z^2 + 2z + (1 − 4) = (z + 1)^2 + 4. This has the roots z = − 1 ± i2. The origin thereby is a spiral sink and is asymptotically stable. The orbits go clockwise around the origin. (c) d dt
x 1 x 2
x 1 x 2
Solution: The characteristic polynomial of the coefficient matrix is p(z) = z^2 + (−1 + 10) = z^2 + 9. This has the roots z = ±i3. The origin thereby is a center and is sta- ble. (It is not asymptotically stable!) The orbits go counterclockwise around the origin.
(8) Suppose you know that for some nonlinear system of differential equations
- the equilibrium solutions are (1,1), (-2,0), and (-2,3);
- for (1,1) the linearization has eigenvalues 3 and 2 with respective eigen- vectors (^) ( 1 1
and
- for (-2,0) the linearization has eigenvalues -1 and -3 with respective eigenvectors ( 1 0
and
- for (-2,3) the linearization has eigenvalues -2 and 1 with respective eigenvectors ( 1 0
and
Sketch a plausible phase portrait for the system. Identify the type and stability of each equilibrium solution. Solution: The type and stability of each equilibrium solution is determined as follows.
- The equilibrium solution (1,1) has two positive simple real eigenvalues. It is thereby a nodal source and is unstable.
- The equilibrium solution (-2,0) has two negative simple real eigenval- ues. It is thereby a nodal sink and is asymptotically stable.
- The equilibrium solution (-2,3) has one negative and one positive sim- ple real eigenvalue. It is thereby a saddle and is unstable. The phase portrait will be sketched during the review.
