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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