Question

Kremer methoed se solve kre
X+2y+z=5
2X+2y+z =6
X+2y+3z=9

Answer 1

$\begin{cases} x+2y+z=5 \\ 2x+2y+z=6 \\ x+2y+3z=9 \end{cases}$

$\Delta=\begin{vmatrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 2 & 3 \end{vmatrix}= \begin{vmatrix} 2 & 1\\ 2 & 3 \end{vmatrix}-2 \begin{vmatrix} 2 & 1\\ 1 & 3 \end{vmatrix}+ \begin{vmatrix} 2 & 2\\ 1 & 2 \end{vmatrix}=4-10+2=-4$

$\Delta_x=\begin{vmatrix} 5 & 2 & 1 \\ 6 & 2 & 1 \\ 9 & 2 & 3 \end{vmatrix}=5 \begin{vmatrix} 2 & 1\\ 2 & 3 \end{vmatrix}-2 \begin{vmatrix} 6 & 1\\ 9 & 3 \end{vmatrix}+ \begin{vmatrix} 6 & 2\\ 9 & 2 \end{vmatrix}=20-18-6=-4$

$\Delta_y=\begin{vmatrix} 1 & 5 & 1 \\ 2 & 6 & 1 \\ 1 & 9 & 3 \end{vmatrix}= \begin{vmatrix} 6 & 1\\ 9 & 3 \end{vmatrix}-5 \begin{vmatrix} 2 & 1\\ 1 & 3 \end{vmatrix}+ \begin{vmatrix} 2 & 6\\ 1 & 9 \end{vmatrix}=9-25+12=-4$

$\Delta_z=\begin{vmatrix} 1 & 2 & 5 \\ 2 & 2 & 6 \\ 1 & 2 & 9 \end{vmatrix}= \begin{vmatrix} 2 & 6\\ 2 & 9 \end{vmatrix}-2 \begin{vmatrix} 2 & 6\\ 1 & 9 \end{vmatrix}+5 \begin{vmatrix} 2 & 2\\ 1 & 2 \end{vmatrix}=6-24+10=-8$

$x=\frac{\Delta_x}{\Delta}=\frac{-4}{-4}=1$

$y=\frac{\Delta_y}{\Delta}=\frac{-4}{-4}=1$

$z=\frac{\Delta_z}{\Delta}=\frac{-8}{-4}=2$