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# Math 0250 Homework 6

• 2.2 Number 1

and the characteristic polynomial is

so the eigenvalues are 5 and -1.

For the eigenvalue 5, the coefficient matrix for the eigenvector equation is

which row reduces to

so an eigenvector for the eigenvalue 5 is

For the eigenvalue -1, the coefficient matrix for the eigenvector equation is

which row reduces to

so an eigenvector for the eigenvalue -1 is

• 2.2 Number 7

and the characteristic polynomial is

Setting this equal to 0 and solving for gives to two eigenvalues .

For the eigenvalue 2+2i, the coefficient matrix for the eigenvector equation is

which row reduces to

An eigenvector for the eigenvalue 2+2i is

We still need to find an eigenvector for the other eigenvalue 2-2i. Here's a trick to find it quickly. In general, if A is a real matrix and U is a complex eigenvector with eigenvalue , then is an eigenvector with eigenvalue . That means that, for the problem at hand, you can find an eigenvector for the eigenvalue 2-2i by just conjugating each coordinate of the one you already found for the eigenvalue 2+2i. Thus the required eigenvalue for 2-2i is just

• 2.2 Number 11

You could calculate the determinant by a cofactor expansion along the first row or last column, but you can save yourself a little effort by first subtracting times the second row from the third. Here's the calculation

so the eigenvalues are 0, 1, and 4.

For the eigenvalue 0, the coefficient matrix of the eigenvector equation is

which row reduces to

Writing out the corresponding equations, and setting the third coordinate equal to, say, 1, gives the eigenvector

For the eigenvalue 1, the coefficient matrix of the eigenvector equation is

which row reduces to

Writing out the corresponding equations, and setting the third coordinate equal to, say, 1, gives the eigenvector

For the eigenvalue 4, the coefficient matrix of the eigenvector equation is

which row reduces to

Writing out the corresponding equations, and setting the third coordinate equal to, say, 1, gives the eigenvector

• 2.3 Number 1
For all the given initial vectors you're going to need the general solution to the equation, which you can generate from the eigenvalues and and eigenvectors. The eigenvalues are 0 and 3, and corresponding eigenvectors are

You can get these by using a computer or calculator, or by hand calculation. The general solution to the differential equation is

Therefore,

so to solve the equation with initial condition X(0)=B, you'll have to solve the algebraic system

Since you're going to have to solve this system for a bunch of different initial conditions, the most efficient way to proceed is to invert the coefficient matrix. Using either row reduction or a computer, the inverse is

To get the values of the constants c1 and c2 for each of the given initial conditions, just multiply each initial vector by A-1.

a

b

c

d

e

f

• 2.3 Number 3
The eigenvectors are 0 and 2, with corresponding eigenvectors

so the general solution to the differential equation is

with initial value

The inverse of the coefficient matrix for the initial condition is

Here are the values of c1 and c2 corresponding to the given initial conditions.

a

b

c

d

e

f

• 2.3 Number 9
The eigenvalues are 2, 4, and 6, with corresponding eigenvectors

so the general solution to the differential equation is

with initial value

The inverse of the coefficient matrix for the initial condition is

Here are the values of c1, c2, and c3 corresponding to the given initial conditions.

a

b

c

d

e

f

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Frank Beatrous
1998-10-25