ExampleOfCramersRule 2002-07-15 22:56:38 2005-05-23 22:46:49 Example Cramer's rule \usepackage{graphicx} %\usepackage{xypic} \usepackage{amsmath} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand{\mathbb}[1]{\mathbbmss{#1}} \newcommand{\figura}[1]{\begin{center}\includegraphics{#1}\end{center}} \newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}} Say we want to solve the system \begin{eqnarray*} 3x + 2y + z -2w&=&4\\ 2x - y + 2z - 5w &=& 15\\ 4x + 2y \phantom{+ 2x} - 5w&=&1\\ 3x \phantom{+ 2y} - 2z -4w &=& 1. \end{eqnarray*} The associated matrix is $$\begin{pmatrix} 3 & 2 & 1 & -2\\ 2 & -1 & 2 & -5\\ 4 & 2 & 0 & -1 \\ 3 & 0 & -2 & -4 \end{pmatrix}$$ whose determinant is $\Delta=-65$. Since the determinant is non-zero, we can use Cramer's rule. To obtain the value of the $k$-th variable, we replace the $k$-th column of the matrix above by the column vector $$\begin{pmatrix} 4\\ 15\\ 1\\ 1\end{pmatrix},$$ the determinant of the obtained matrix is divided by $\Delta$ and the resulting value is the wanted solution. So $$ x=\frac{\Delta_1}{\Delta}= \frac{ \begin{vmatrix} 4 & 2 & 1 & -2\\ 15 & -1 & 2 & -5\\ 1 & 2 & 0 & -1\\ 1 & 0 & -2 & -4 \end{vmatrix} }{-65}=\frac{-65}{-65}=1 $$ $$ y=\frac{\Delta_2}{\Delta}= \frac{ \begin{vmatrix} 3 & 4 & 1 & -2\\ 2 & 15 & 2 & -5\\ 4 & 1 & 0 & -1\\ 3 & 1 & -2 & -4 \end{vmatrix} }{-65}=\frac{130}{-65}=2 $$ $$ z=\frac{\Delta_3}{\Delta}= \frac{ \begin{vmatrix} 3 & 2 & 4 & -2\\ 2 & -1 & 15 & -5\\ 4 & 2 & 1 & 1\\ 3 & 0 & 1 & -4 \end{vmatrix} }{-65}=\frac{-195}{-65}=3 $$ $$ w=\frac{\Delta_4}{\Delta}= \frac{ \begin{vmatrix} 3 & 2 & 1& 4 \\ 2 & -1 & 2 & 15\\ 4 & 2 & 0 & 1 \\ 3 & 0 & -2 & 1 \end{vmatrix} }{-65}=\frac{65}{-65}=-1 $$