Sunday, July 20, 2014

Inverse of a 2x2 Matrix


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In my previous post (http://mikemstech.blogspot.com/2014/07/c-matrix-inversion-with-latex-output.html) I demonstrated an application that can generate the steps to show the inversion of a matrix by Gauss Jordan elimination.

In a few posts, I plan to answer the following questions:
What is the inverse of a 1x1 Matrix?
What is the inverse of a 2x2 Matrix?
What is the inverse of a 3x3 Matrix?
What is the inverse of a 4x4 Matrix?

Back to Mike's Big Data, Data Mining, and Analytics Tutorial

The inverse of a 2x2 matrix is defined as follows. For a 2x2 matrix:

$$ A = \begin{pmatrix}a & b \\ c & d\end{pmatrix} $$

Unsimplified, the inverse is equal to:

$$ A^{-1} = \begin{pmatrix}\frac{1}{a} - \left (\frac{b}{a} \right )  \left ( \frac{0 - \left (c \right ) \left ( \frac{1}{a}\right )}{d - \left (c \right ) \left ( \frac{b}{a}\right )}\right )
&
0 - \left (\frac{b}{a} \right )  \left ( \frac{1}{d - \left (c \right ) \left ( \frac{b}{a}\right )}\right )
\\\frac{0 - \left (c \right ) \left ( \frac{1}{a}\right )}{d - \left (c \right ) \left ( \frac{b}{a}\right )}
&
\frac{1}{d - \left (c \right ) \left ( \frac{b}{a}\right )}
\end{pmatrix} $$

After simplifying, this is the inverse of a 2x2 matrix.

$$ A^{-1} = \frac{1}{ad-bc} \begin{pmatrix}  d & -b \\ -c & a  \end {pmatrix} $$

The latex code generated for a 2x2 inverse is the following (note that I changed the default variables to a, b, c, and d to match how most people learn the 2x2 matrix):

\documentclass{article}

% This is the output from LatexMatrixInverse 1.0 for a matrix with rank 2 
% For more information on this application, see
% http://mikemstech.blogspot.com

\usepackage{geometry}

% Note: you should probably use pdflatex to compiile this file. 
% Other processors are known to have some issues with using
% 'geometry' to set paper size

% Adjust the page size here if output is wrapping in a bad way.
% Default is 8.5 x 11 in (Letter)
\geometry{papersize={16in,11in}}

%Import AMS Latex packages
\usepackage{amsmath, amssymb}
\setcounter{MaxMatrixCols}{5}


%Variable definition

\begin{document}
% Definition of initial A
% A row 1
\newcommand{\ARbCb}{a}
\newcommand{\ARbCc}{b}

% A row 2
\newcommand{\ARcCb}{c}
\newcommand{\ARcCc}{d}

% Definition of initial B
\newcommand{\BRb}{e}
\newcommand{\BRc}{f}



LatexMatrixInverse 1.0 Output for rank 2, ShowIntermediateSteps is True.

For more information on this application, please see http://mikemstech.blogspot.com

Given the following initial matrices:
\begin{equation*}
A = \begin{pmatrix}\ARbCb
&\ARbCc
\\\ARcCb
&\ARcCc
\end{pmatrix}B = \begin{pmatrix}\BRb\\ \BRc\end{pmatrix}\end{equation*}
 We want to find $A^{-1}$ and $A^{-1} B$...
\begin{equation*}
\left ( \begin{array}{cc|cc|c}\ARbCb
&\ARbCc
&1
&0
&\BRb\\
\ARcCb
&\ARcCc
&0
&1
&\BRc\end{array} \right )
\end{equation*}
\begin{equation*}
\xrightarrow{\frac{1}{\ARbCb} R_{1}}
\left ( \begin{array}{cc|cc|c}1
&\frac{\ARbCc}{\ARbCb}
&\frac{1}{\ARbCb}
&0
&\frac{\BRb}{\ARbCb}\\
\ARcCb
&\ARcCc
&0
&1
&\BRc\end{array} \right )
\end{equation*}
\begin{equation*}
\xrightarrow{R_{2} - \left ( \ARcCb\right ) R_{1}}
\left ( \begin{array}{cc|cc|c}1
&\frac{\ARbCc}{\ARbCb}
&\frac{1}{\ARbCb}
&0
&\frac{\BRb}{\ARbCb}\\
0
&\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )
&0 - \left (\ARcCb \right ) \left ( \frac{1}{\ARbCb}\right )
&1
&\BRc - \left (\ARcCb \right ) \left ( \frac{\BRb}{\ARbCb}\right )\end{array} \right )
\end{equation*}
\begin{equation*}
\xrightarrow{\frac{1}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )} R_{2}}
\left ( \begin{array}{cc|cc|c}1
&\frac{\ARbCc}{\ARbCb}
&\frac{1}{\ARbCb}
&0
&\frac{\BRb}{\ARbCb}\\
0
&1
&\frac{0 - \left (\ARcCb \right ) \left ( \frac{1}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
&\frac{1}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
&\frac{\BRc - \left (\ARcCb \right ) \left ( \frac{\BRb}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\end{array} \right )
\end{equation*}
\begin{equation*}
\xrightarrow{R_{1} - \left ( \frac{\ARbCc}{\ARbCb}\right ) R_{2}}
\left ( \begin{array}{cc|cc|c}1
&0
&\frac{1}{\ARbCb} - \left (\frac{\ARbCc}{\ARbCb} \right ) \linebreak \left ( \frac{0 - \left (\ARcCb \right ) 
\left ( \frac{1}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\right )
&0 - \left (\frac{\ARbCc}{\ARbCb} \right ) \linebreak \left ( \frac{1}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\right )
&\frac{\BRb}{\ARbCb} - \left (\frac{\ARbCc}{\ARbCb} \right ) \linebreak \left ( \frac{\BRc - \left (\ARcCb \right ) 
\left ( \frac{\BRb}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\right )\\
0
&1
&\frac{0 - \left (\ARcCb \right ) \left ( \frac{1}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
&\frac{1}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
&\frac{\BRc - \left (\ARcCb \right ) \left ( \frac{\BRb}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\end{array} \right )
\end{equation*}
\begin{equation*}
A^{-1} = \begin{pmatrix}\frac{1}{\ARbCb} - \left (\frac{\ARbCc}{\ARbCb} \right ) \linebreak \left ( \frac{0 - \left (\ARcCb \right ) 
\left ( \frac{1}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\right )
&
0 - \left (\frac{\ARbCc}{\ARbCb} \right ) \linebreak \left ( \frac{1}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\right )
\\\frac{0 - \left (\ARcCb \right ) \left ( \frac{1}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
&
\frac{1}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
\end{pmatrix}A^{-1}B = \begin{pmatrix}\frac{\BRb}{\ARbCb} - \left (\frac{\ARbCc}{\ARbCb} \right ) \linebreak \left ( \frac{\BRc - \left (\ARcCb \right ) 
\left ( \frac{\BRb}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}\right )\\ \frac{\BRc - \left (\ARcCb \right ) 
\left ( \frac{\BRb}{\ARbCb}\right )}{\ARcCc - \left (\ARcCb \right ) \left ( \frac{\ARbCc}{\ARbCb}\right )}
\end{pmatrix}\end{equation*}
\end{document}


Here is a screenshot of the work for calculating a 2x2 inverse (based on the compiled code above):































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