| Line 61: | Line 61: | ||
<br> | <br> | ||
| − | This then allows us to write: | + | This then allows us to write: |
| + | <math>A{{A}^{-1}}={{A}^{-1}}A={{I}_{n}}</math> | ||
| − | |||
| − | |||
| − | |||
---- | ---- | ||
Revision as of 11:56, 12 December 2010
The Inverse of a Matrix
In linear algebra, the study of matrices is one of the fundamental basis of this subject. One of the concepts within this study, is the notion of an invertible or nonsingular matrix.
Definition
A square matrix is said to be invertible or nonsingular, if when multiplied by another similar matrix, the result yields the identity matrix.
Let A and B be n × n matrices and In be the n × n identity matrix
A is invertible or nonsingular and B is its inverse if:
$ \begin{align} & AB=BA={{I}_{n}} \\ & \\ & \overbrace{\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & \cdots & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & \cdots & {{a}_{2n}} \\ \vdots & \vdots & \ddots & \vdots \\ {{a}_{n1}} & {{a}_{n2}} & \cdots & {{a}_{nn}} \\ \end{matrix} \right)}^{A}\overbrace{\left( \begin{matrix} {{b}_{11}} & {{b}_{12}} & \cdots & {{b}_{1n}} \\ {{b}_{21}} & {{b}_{22}} & \cdots & {{b}_{2n}} \\ \vdots & \vdots & \ddots & \vdots \\ {{b}_{n1}} & {{b}_{n2}} & \cdots & {{b}_{nn}} \\ \end{matrix} \right)}^{B}=\overbrace{\left( \begin{matrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{matrix} \right)}^{{{I}_{n}}} \\ \end{align} $
If the above condition is not met, and the determinant of matrix A is 0, then it is called singular or noninvertible.
If matrix B is matrix A's inverse, then it is written as A-1. Matrix B is a unique matrix; there only exists one or no possible inverse of matrix A. The proof of this notion is summarized below:
$ \begin{align} & \text{Let }B\text{ and }C\text{ be inverses of }A\text{. Then:} \\ & \\ & AB=BA={{I}_{n}}\text{ and }AC=CA={{I}_{n}} \\ & \\ & \text{Since any matrix multiplied by the identity matrix of the same dimension yields that same matrix, then:} \\ & \\ & B=B{{I}_{n}}=B(AC)=(BA)C={{I}_{n}}C=C \\ & \Rightarrow B=C\text{ and thus the inverse matrix, if it exists, is unique}\text{.} \\ \end{align} $
This then allows us to write:
$ A{{A}^{-1}}={{A}^{-1}}A={{I}_{n}} $
