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<br> | <br> | ||
− | + | <math>\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 \\ | + | 1 & 0 & \cdots & 0 \\ |
− | 0 & 1 & \cdots & 0 \\ | + | 0 & 1 & \cdots & 0 \\ |
− | \vdots & \vdots & \ddots & \vdots \\ | + | \vdots & \vdots & \ddots & \vdots \\ |
− | 0 & 0 & \cdots & 1 \end{ | + | 0 & 0 & \cdots & 1 \\ |
− | </math> | + | \end{matrix} \right)}^{{{I}_{n}}} \\ |
+ | \end{align}</math> | ||
[[Category:MA265Fall2010Walther]] | [[Category:MA265Fall2010Walther]] |
Revision as of 14:22, 27 November 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.
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} $