15.3. AUTOMATION WITH MATLAB 355

This is only a 3×3 matrix and so it is not hard to estimate the eigenvalues. Just getthe characteristic equation, graph it using a calculator and zoom in to find the eigenvalues.If you do this, you find there is an eigenvalue near −1.2, one near −.4, and one near 5.5.(The characteristic equation is 2+ 8λ + 4λ

2−λ3 = 0.) Of course we have no idea what

the eigenvectors are.Lets first try to find the eigenvector and a better approximation for the eigenvalue near

−1.2. In this case, let α =−1.2. Then

(A−αI)−1 =

 −25.357143 −33.928571 50.012.5 17.5 −25.0

23.214286 30.357143 −45.0

 .

Then  −25.357143 −33.928571 50.012.5 17.5 −25.0

23.214286 30.357143 −45.0

17 1

11

=

 −4.9432×1028

2.4312×1028

4.4928×1028

The initial approximation for an eigenvector will then be the above divided by its largestentry.  −4.9432×1028

2.4312×1028

4.4928×1028

 1−4.9432×1028 =

 1.0−0.49183−0.90888

How close is this to being an eigenvector? 2 1 3

2 1 13 2 1

 1.0−0.49183−0.90888

=

 −1.21850.599291.1075



−1.2185

 1.0−0.49183−0.90888

=

 −1.21850.599291.1075

For all practical purposes, this has found the eigenvector for the eigenvalue −1.2185.

Next we shall find the eigenvector and a more precise value for the eigenvalue near−.4.In this case,

(A−αI)−1 =

 8.0645161×10−2 −9.2741935 6.4516129−.40322581 11.370968 −7.2580645.40322581 3.6290323 −2.7419355

 .

The first approximation to an eigenvector can be obtained as before.