*2.6 Matrix Inversion

Computing the inverse of a matrix and solving simultaneous equations are related tasks. The most economical way to invert an n n matrix A is to solve the equations

(2.33)

where I is the n n identity matrix. The solution X, also of size n n, will be the inverse of A. The proof is simple: after we premultiply both sides of Eq. (2.33) by A ¹ we have A ¹ AX= A ¹ I, which reduces to X= A ^1.

Inversion of large matrices should be avoided whenever possible due its high cost. As seen from Eq. (2.33), inversion of A is equivalent to solving Ax _i= b _i, i=1, 2,..., n, where b _i is the ith column of I. If LU decomposition is employed in the solution, the solution phase (forward and back substitution) must be repeated n times, once for each b _i. Since the cost of computation is proportional to n ³ for the decomposition phase and n ² for each vector of the solution phase, the cost of inversion is considerably more expensive than the solution of Ax= b (single constant vector b).

Matrix inversion has another serious drawback a banded matrix loses its structure during inversion. In other words, if A is banded or otherwise sparse, then A ^?1 is fully...