Cramerovo pravilo

Cramerovim pravilom riješite sustav

$\displaystyle \begin{matrix} 2x_1 & + & x_{2} & + & x_{3} & = & 2, \\ x_{1}& +... ...2} & + & x_{3} & = & 3, \\ x_{1} & + & x_{2} & + & 2x_3 & = & -1. \end{matrix}$

Rješenje. Matrica sustava je kvadratna i regularna jer je

$\displaystyle \det A=\begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 1\\ 1 & 1 & 2 \end{vmatrix}=8+1+1-2-2-2=4\neq0.$

Stoga prema

[M1, teorem 2.10] vrijedi

$\displaystyle x_i=\frac{D_i}{\det A},\quad i=1,2,3,$

gdje je

$\displaystyle D_1$	$\displaystyle =\begin{vmatrix}2 & 1 & 1 \\ 3 & 2 & 1\\ -1 & 1 & 2 \end{vmatrix} =8-1+3+2-2-6=4,$
$\displaystyle D_2$	$\displaystyle =\begin{vmatrix}2 & 2 & 1 \\ 1 & 3 & 1\\ 1 & -1 & 2 \end{vmatrix} =12+2-1-3+2-4=8,$
$\displaystyle D_3$	$\displaystyle =\begin{vmatrix}2 & 1 & 2 \\ 1 & 2 & 3\\ 1 & 1 & -1 \end{vmatrix} =-4+3+2-4-6+1=-8.$

Slijedi

$\displaystyle x_1=\frac{4}{4},\quad x_2=\frac{8}{4},\quad x_3=\frac{-8}{4},$

pa rješnje sustava glasi

$\displaystyle \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 1\\ 2\\ -2 \end{bmatrix}.$