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Example 8.10
Using the Cayley-Hamilton theorem, find the inverse of the matrix M given
in Example 8.9.

Solution: The characteristic equation for this matrix is given by:

2
p(M) = M – 5M + 4I = 0

–1
Now multiply this equation by M to obtain:
–1
M – 5I + 4M = 0

and

 3 
−1 
⇒ M −1 = 025 5 I − M) =  4
.(
 1 1 
 − 
 8 2 
Example 8.11
Reduce the following fourth-order polynomial in M, where M is given in
Example 8.9, to a first-order polynomial in M:

3
2
4
P(M) = M + M + M + M + I
Solution: From the results of Example 8.10 , we have:

M = 5 M 4− I
2
M = 5 M − 4 M = 5 5( M 4− I −) 4 M = 21 M 20− I
3
2
M = 21 M − 20 M = 21 5( M 4− I ) − 20 M = 85 M 84− I
2
4
⇒ (P M =) 112 M 107− I

Verify the answer numerically using MATLAB.

dX
8.9.2 Solution of Equations of the Form = AX
dt
We sketched a technique in Pb. 8.17 that uses the eigenvectors matrix and
solves this equation. In Example 8.12, we solve the same problem using the
Cayley-Hamilton technique.

© 2001 by CRC Press LLC