Matrices and Tensors - Continuum Mechanics of Anisotropic Materials - page 344

Example A.5.4

Find the inverse of the matrix

2 3

18 6 6
6 50
601

4 5

A ¼

Solution: The matrix of cofactors is given by

2 3

315 126 90

4 5 ;

co A ¼ 126 342 36

90 36 234

thus the inverse of A is then given by

2 3

17 : 5 7 5

co A T
Det A ¼

1
243

4 5 :

A 1

¼ 792

523

t addresses the
question of the n-tuple t being transformed by A into some scalar multiple of itself,
l

The eigenvalue problem for a linear transformation r ¼ A

and
t exist, they are called eigenvalues and eigen n-tuples of the matrix A , respectively.
The eigenvalue problem is then to find solutions to the equation

t . Specifically, for what values of t and l does l t ¼ A t ? If such values of l

ð A l 1 Þ t ¼ :

0 (A.48)

This is a system of linear equations for the elements of the n-tuple t . For the case
of n ¼ 3 it may be written in the form:

ð A 11 lÞ t 1 þ A 12 t 2 þ A 13 t 3 ¼ 0 ;

A 21 t 1 þð A 22 lÞ t 2 þ A 23 t 3 ¼ 0 ;

A 31 t 1 þ A 32 t 2 þð A 33 lÞ t 3 ¼ 0 : (A.49)

The standard approach to the solution of a system of linear equations like (A.48)
is Cramer's rule. For a system of three equations in three unknowns, (A.36) with
n

¼ 3,

r 1 ¼ A 11 t 1 þ A 12 t 2 þ A 13 t 3 ;

r 2 ¼ A 21 t 1 þ A 22 t 2 þ A 23 t 3 ;

r 3 ¼ A 31 t 1 þ A 32 t 2 þ A 33 t 3 ; (A.50)

Cramer's rule provides the solution for the n-tuple t ¼ [ t 1 , t 2 , t 3 ]: