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Mathematics-Online course: Linear Algebra - Matrices - Matrix Operations

Inverse Matrix

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For an invertible linear map $ v\mapsto Av$ the inverse of matrix $ A$ is denoted by $ A^{-1}$, that is,

$\displaystyle A A^{-1} = A^{-1} A = E $

where

$\displaystyle E = \left(\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \c... ... \\ \vdots & 0 & \ddots & \vdots \\ 0 & & \cdots & 1 \end{array}\right) $

is the identity matrix.
(Authors: Burkhardt/Höllig/Hörner)
The invertible (or regular) matrices $ A\in K^{n\times n}$ form a group with respect to the matrix multiplication. This group is called general linear group and is denoted by $ \operatorname{GL}(n,K)$.

We have

$\displaystyle (AB)^{-1} = B^{-1} A^{-1} $

for $ A,B\in \operatorname{GL}(n,K)$.
We have:

$\displaystyle B^{-1}A^{-1}AB = B^{-1}EB=B^{-1}B=E $

Thus, $ B^{-1}A^{-1}$ is the inverse of $ AB$.

(Authors: Burkhardt/Höllig/Hörner)
We obtain the inverse of a diagonal matrix by replacing each diagonal entry with its reciprocal value. For example:

$\displaystyle A= \left(\begin{array}{rr} 3 & 0 \\ 0 & 1/7 \end{array}\right)... ... A^{-1} = \left(\begin{array}{rr} 1/3 & 0 \\ 0 & 7 \end{array}\right)\,. $

The inverse of an upper (lower) triangle matrix is an upper (lower) triangle matrix:

$\displaystyle B= \left(\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array}\right)\,... ... B^{-1} = \left(\begin{array}{rr} 1 & 0 \\ -1 & 1 \end{array}\right)\,. $

However, in many cases the inverse of a matrix contains only non-zero entries:

$\displaystyle C= \left(\begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array}\right)... ...C^{-1} = \left(\begin{array}{rr} 1 & -2 \\ 1 & -1 \end{array}\right) \,. $

The entries have to be found by solving a linear system of equations.

(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005