problems on solving systems with the eigenvalue method

7/29/2019 Problems on Solving Systems With the Eigenvalue Method

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MAP2302 (summer 2012)

Addit ional Problems Involv ing Systems and Eigenvalues (w ith solu tions)

Problem 1 Find the eigenvalues and eigenvectors of the matrix 3 21 0

Problem 2 Solve the system 2 4 3

Problem 3 Solve the system

149

Problem 4 (system with 0 as an eigenvalue) Solve the system 3 9 3


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Solutions

Solution Problem 1

The eigenvalues are found from the condition

det 3 21 0 0which gives us the characteristic equation

3 0 1 2 0that is,

3 2 0The eigenvalues are therefore 2 and 1. To find the corresponding eigenvectors, we solve theequation For :

3 21 0

00

That is

1 21 2 00This gives rise to the system

2 0 2 0

Notice that the second equation is a multiple of the first, that is, the system is redundant. This is why ithas infinitely many solutions (besides the trivial one).

Solving for thepivot variable, 2Thus,

2

2

1. Since is arbitrary, we can set it equal to 1. Thus,

21So in fact we have found infinitely may eigenvectors corresponding to 2. We just need one and thetraditional letter used to denote eigenvectors is the Greek letter xi ;Hence, let ususe 21 as the eigenvector associated with the eigenvalue 2.For ,

3 21 0

00

That is

2 21 1

00

This gives rise to the system 2 2 0 0

and again the equations are multiples of each other.

Solving for the pivot variable,


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Thus,

1

1. Again, we have an infinite number of eigenvalues, all parallel oranti-parallel to each other. We may set 1 to obtain

11as the eigenvector corresponding to 1.

Thus, we have two eigenevalues: 2 and 1 and their corresponding eigenvectors 21 and

11

Solution Problem 2

The matrix of coefficients is 1 24 3.Step 1: Find the eigenvalues of the matrix of coefficients by solving

1 24 3 0The characteristic equation is

1 3 8 0

4 3 8 0 4 5 0

Therefore, the eigenvalues of the matrix are 5 1.Step 2: Find the corresponding eigenvectors by substituting each eigenvalue into the equation

00

For 5:1 5 24 3 5

00

NOTE1

: see below for remarks about notation.Thus, the first eigenvector must have components that satisfy

4 24 2

00

which gives rise to the system4 2 0

4 2 0This system has infinitely many solutions because one equation is a multiple of the other (the secondequation is 1 the first equation) so we can pick either equation to find the eigenvector and solve foreither component. To avoid fractions, let us solve for:

2

Thus, the eigenvectors corresponding to 5 are given by

1If we really wanted to be precise, we would write ,, for the eigenvector corresponding to and ,, for the eigenvector corresponding to .


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2 12

We can let 1 to obtain 12 as the eigenvector corresponding to 5.For 1:

1 1 24 3 1 00Thus

2 24 4

00

The corresponding system is2 2 04 4 0

Once again, we have a system with infinitely many solutions given by the relation

0 Thus,

1

1. Set

1to obtain

1

1.

Step 3: We have two linearly independent solutions:

12

and

11

Step 4: Using the principle of superposition, write the general solution in vector form:

Thus,

12

1

1

is oursolution vector.

Solution problem 3

The matrix of coefficients is 0 114 9.Step 1: Find the eigenvalues of the matrix of coefficients by solving

0 114 9 0The characteristic polynomial is

9 1 4 0

9 1 4 0whose roots are 7 2 so we have the eigenvalues.Step 2: Find the corresponding eigenvectors by substituting each eigenvalue into the equation

0 114 9

00

For 7:0 7 114 9 7

00


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Thus,

7 114 2

00

which gives rise to the system7 0

14 2 0The second equation is2 the first equation. We have

7The eigenvectors corresponding to 7 are given by

7 1

7

Thus, we can take 17For 2:

0 2 114 9 2

00

we obtain

2 114 7 00The corresponding system is

2 014 7 0

The second equation being7 the first so we really have only one equation, as is expected.2 0 2

Thus,

2 1

2 and we can take 1

2.Step 3: The two linearly independent solutions are

17 and

12

Step 4: Write the solution in vector form using the superposition principle:

Thus,

17 12

In component form,

7 2

Solution Problem 4

The characteristic polynomial is 6. Therefore, the matrixs eigenvalues are 6 , 0.Eigenvectors for :2

2 Any other letters can be used besides and to denote the components of the eigenvectors.


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3 0 91 3 0

00

Thus,3 9 0

3 0Using the second equation,

3; arbitraryso we may let

1to obtain the eigenvector

31To find the eigenvector corresponding to , we solve

3 6 91 3 6

00

or

3 9 0 3 0

The second equation is3 the first equation. Hence, the infinite solutions are given by either:For example,

3 0or

3 . Thus,

3 31

And we may take 31The solution vector is given by

problems on solving systems with the eigenvalue method

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