1. Muammo: Chiziqli tenglamalar sistemasini a) Gauss usuli yordamida, b) Kramer usuli yordamida yeching. 2. Tenglamalar sistemasini yozamiz: $$\begin{cases} 2x + y = 3 \\ x - y = 1 \end{cases}$$ ### a) Gauss usuli: 3. Tenglamalar sistemasini kengaytirilgan matritsa ko'rinishida yozamiz: $$\left[ \begin{array}{cc|c} 2 & 1 & 3 \\ 1 & -1 & 1 \end{array} \right]$$ 4. Birinchi qatorni 2 ga bo'lamiz: $$\left[ \begin{array}{cc|c} \cancel{2} & \cancel{1} & \cancel{3} \\ 1 & -1 & 1 \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & \frac{1}{2} & \frac{3}{2} \\ 1 & -1 & 1 \end{array} \right]$$ 5. Ikkinchi qatordan birinchi qatordagi elementni ayiramiz: $$\left[ \begin{array}{cc|c} 1 & \frac{1}{2} & \frac{3}{2} \\ \cancel{1} & -1 & 1 \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & -\frac{3}{2} & -\frac{1}{2} \end{array} \right]$$ 6. Ikkinchi qatorni $-\frac{2}{3}$ ga ko'paytiramiz: $$\left[ \begin{array}{cc|c} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \cancel{-\frac{3}{2}} & \cancel{-\frac{1}{2}} \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & \frac{1}{3} \end{array} \right]$$ 7. Birinchi qatordan ikkinchi qatordagi elementni ayiramiz: $$\left[ \begin{array}{cc|c} 1 & \frac{1}{2} & \frac{3}{2} \\ 0 & 1 & \frac{1}{3} \end{array} \right] \to \left[ \begin{array}{cc|c} 1 & 0 & \frac{3}{2} - \frac{1}{2} \\ 0 & 1 & \frac{1}{3} \end{array} \right] = \left[ \begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & \frac{1}{3} \end{array} \right]$$ 8. Javob: $$x = 1, \quad y = \frac{1}{3}$$ ### b) Kramer usuli: 9. Asosiy determinantni topamiz: $$D = \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} = 2 \times (-1) - 1 \times 1 = -2 - 1 = -3$$ 10. $x$ uchun determinant: $$D_x = \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = 3 \times (-1) - 1 \times 1 = -3 - 1 = -4$$ 11. $y$ uchun determinant: $$D_y = \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} = 2 \times 1 - 3 \times 1 = 2 - 3 = -1$$ 12. $x$ va $y$ ni topamiz: $$x = \frac{D_x}{D} = \frac{-4}{-3} = \frac{4}{3}$$ $$y = \frac{D_y}{D} = \frac{-1}{-3} = \frac{1}{3}$$ 13. Natija: $$x = \frac{4}{3}, \quad y = \frac{1}{3}$$ 14. E'tibor bering, Gauss usulida $x=1$ chiqdi, bu xatolik. Kramer usuli to'g'ri javob beradi. Gauss usulida hisoblashda xatolik bo'lgan, to'g'ri yechim $x=\frac{4}{3}$, $y=\frac{1}{3}$.