Matrix Inverses

Question

So in class we have been discussing matrix inverses and the quickest way that I know of is to get a matrix A, and put it side by side with the identity matrix, like $[A|I_{n}]$ and apply the Gauss-Jordan algorithm until it is of the form $[I_{n}|A^{-1}$], where $A^{-1}$ will show up assuming A is invertible.

We also discussed using the formula $A^{-1}=\frac{\operatorname{adj}(A)}{\det(A)}$, however after a few examples, it was clear that this formula would take far too long to find the inverse of A as the matrix size got bigger.

Is the first method I described the quickest way to find a inverse of a matrix or is there a more efficient way?

Answer 1

This depends on whether you're talking about "pen-and-paper" based methods, or computer-based ones.

For pen-and-paper, the Gauss-Jordan elimination is easier and less error-prone than other methods most of the time. In some special cases, the determinants might be easy to calculate because of the matrix having some special form and thus allow the inverse to be found faster -- but for general matrix, it's hard to beat the G-J approach.

On the computer, there are faster methods available (useful mainly for matrices of large dimensions); the fastest ones for general matrices run as fast as matrix multiplication can be performed.

Wikipedia has a nice list of various methods.

Answer 2

It depends on context: There happen to be a number of techniques to use to find the inverse of a matrix, some of which you may not have yet learned. By paper-pencil, the Gauss-Jordan method is probably easiest, and quickest, to use.

See Methods of Matrix Inversion, in the invertible matrix entry of Wikipedia, you'll fine the methods you mention, as well as other methods for finding the inverse of an invertible matrix.

N.B. Recall: not all $n\times n$ matrices have an inverse! (So when you can easily do so, it might be a good idea to find the determinant prior to attempting to find an inverse.)

Answer 3

The fast way to get the inverse of the matrix $A$, is when you note that the solution of linear system $$Ax = e_i,$$ where $e_i = (0,...,0,\overbrace{1}^{i \text{ position}},0,...,0)^T$, is $x = i$ th column of the matriz $A^{-1}$. Then, in order to compute $A^{-1}$, you need to compute all its columns, and for that, you need to solve the $n$ linear systems $$Ax_i\ =\ e_i,\quad i=1,2,...,n$$ So, the velocity of compute $A^{-1}$ depends of the velocity of solve a linear system. When you use Gaussian Elimination or $LU$-Factorization, you get the Gauss-Jordan method, but you can change the solver, but for "pen-and-paper", as @Peter Košinár said, is faster Gauss-Jordan.