
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 22i.
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
22i by just conjugating each coordinate of the one you already found for
the eigenvalue 2+2i. Thus the required eigenvalue for 22i 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 c_{1} and c_{2} 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 c_{1} and c_{2} 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 c_{1}, c_{2}, and c_{3} corresponding to the given
initial conditions.
 a

 b

 c

 d

 e

 f
