example of Cramer’s rule

Say we want to solve the system

$\displaystyle 3x+2y+z-2w$	$\displaystyle=$	$\displaystyle 4$
$\displaystyle 2x-y+2z-5w$	$\displaystyle=$	$\displaystyle 15$
$\displaystyle 4x+2y\phantom{+2x}-5w$	$\displaystyle=$	$\displaystyle 1$
$\displaystyle 3x\phantom{+2y}-2z-4w$	$\displaystyle=$	$\displaystyle 1.$

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.

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

Title	example of Cramer’s rule
Canonical name	ExampleOfCramersRule
Date of creation	2013-03-22 12:50:33
Last modified on	2013-03-22 12:50:33
Owner	drini (3)
Last modified by	drini (3)
Numerical id	7
Author	drini (3)
Entry type	Example
Classification	msc 15A15