1. Muammo: Berilgan tenglama sistemasini yeching: $$\begin{cases} a = x_1 + x_2 + 2x_3 \\ b = -1 \\ X = A^{-1}B \end{cases}$$ va quyidagi tenglamalar: $$\begin{cases} 2x_1 - x_2 + 2x_3 = -4 \\ 4x_1 + x_2 + 4x_3 = -2 \end{cases}$$ 2. Bu yerda $A$ matritsasi va $B$ vektori berilgan, $X = A^{-1}B$ formulasi orqali $X$ ni topamiz. 3. Avvalo, $A$ matritsasini va $B$ vektorini aniqlaymiz: $$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 2 \\ 4 & 1 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} a \\ -4 \\ -2 \end{bmatrix}$$ 4. $a = x_1 + x_2 + 2x_3$ deb berilgan, lekin $a$ ni $X$ orqali ifodalash kerak emas, chunki $a$ tenglama tizimining birinchi qatori bilan mos keladi. 5. Tenglama tizimini quyidagicha yozamiz: $$\begin{cases} x_1 + x_2 + 2x_3 = a \\ 2x_1 - x_2 + 2x_3 = -4 \\ 4x_1 + x_2 + 4x_3 = -2 \end{cases}$$ 6. Bu yerda $a$ va $b$ qiymatlari berilgan emas, faqat $b = -1$ deb ko'rsatilgan, lekin $b$ tenglamalarda ishlatilmagan. Shuning uchun $a$ ni $x_1,x_2,x_3$ orqali ifodalash kerak. 7. $A$ matritsasining determinantini topamiz: $$\det(A) = 1 \times \begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} - 1 \times \begin{vmatrix} 2 & 2 \\ 4 & 4 \end{vmatrix} + 2 \times \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix}$$ $$= 1((-1)(4) - 2(1)) - 1(2 \times 4 - 2 \times 4) + 2(2 \times 1 - (-1) \times 4)$$ $$= 1(-4 - 2) - 1(8 - 8) + 2(2 + 4) = 1(-6) - 0 + 2(6) = -6 + 12 = 6$$ 8. Determinant $6 \neq 0$, shuning uchun $A$ invertible, $X = A^{-1}B$ formulasi ishlaydi. 9. $A^{-1}$ ni topish uchun adjoint va determinantdan foydalanamiz, yoki Cramer qoidasi yordamida $x_1,x_2,x_3$ ni topamiz. 10. Cramer qoidasi bo'yicha: $$x_i = \frac{\det(A_i)}{\det(A)}$$ bu yerda $A_i$ - $A$ matritsasining $i$-ustuni $B$ vektori bilan almashtirilgan matritsa. 11. $B$ vektori: $$B = \begin{bmatrix} a \\ -4 \\ -2 \end{bmatrix}$$ Lekin $a = x_1 + x_2 + 2x_3$ bo'lgani uchun, $a$ ni $x_i$ orqali ifodalash tizimni yopiq holda yechishni qiyinlashtiradi. 12. Shuning uchun, $a$ ni $x_1,x_2,x_3$ o'rniga $a$ deb qoldirib, $x_1,x_2,x_3$ ni $a$ ga bog'liq holda ifodalaymiz. 13. $A_1$ matritsasi: $$A_1 = \begin{bmatrix} a & 1 & 2 \\ -4 & -1 & 2 \\ -2 & 1 & 4 \end{bmatrix}$$ 14. $\det(A_1)$ ni hisoblaymiz: $$\det(A_1) = a \times \begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} - 1 \times \begin{vmatrix} -4 & 2 \\ -2 & 4 \end{vmatrix} + 2 \times \begin{vmatrix} -4 & -1 \\ -2 & 1 \end{vmatrix}$$ $$= a((-1)(4) - 2(1)) - 1((-4)(4) - 2(-2)) + 2((-4)(1) - (-1)(-2))$$ $$= a(-4 - 2) - 1(-16 + 4) + 2(-4 - 2) = -6a + 12 - 12 = -6a$$ 15. $x_1 = \frac{\det(A_1)}{\det(A)} = \frac{-6a}{6} = -a$ 16. $A_2$ matritsasi: $$A_2 = \begin{bmatrix} 1 & a & 2 \\ 2 & -4 & 2 \\ 4 & -2 & 4 \end{bmatrix}$$ 17. $\det(A_2)$ ni hisoblaymiz: $$\det(A_2) = 1 \times \begin{vmatrix} -4 & 2 \\ -2 & 4 \end{vmatrix} - a \times \begin{vmatrix} 2 & 2 \\ 4 & 4 \end{vmatrix} + 2 \times \begin{vmatrix} 2 & -4 \\ 4 & -2 \end{vmatrix}$$ $$= 1((-4)(4) - 2(-2)) - a(2 \times 4 - 2 \times 4) + 2(2 \times -2 - (-4) \times 4)$$ $$= 1(-16 + 4) - a(8 - 8) + 2(-4 + 16) = -12 + 0 + 24 = 12$$ 18. $x_2 = \frac{\det(A_2)}{\det(A)} = \frac{12}{6} = 2$ 19. $A_3$ matritsasi: $$A_3 = \begin{bmatrix} 1 & 1 & a \\ 2 & -1 & -4 \\ 4 & 1 & -2 \end{bmatrix}$$ 20. $\det(A_3)$ ni hisoblaymiz: $$\det(A_3) = 1 \times \begin{vmatrix} -1 & -4 \\ 1 & -2 \end{vmatrix} - 1 \times \begin{vmatrix} 2 & -4 \\ 4 & -2 \end{vmatrix} + a \times \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix}$$ $$= 1((-1)(-2) - (-4)(1)) - 1(2(-2) - (-4)(4)) + a(2 \times 1 - (-1) \times 4)$$ $$= 1(2 + 4) - 1(-4 + 16) + a(2 + 4) = 6 - 12 + 6a = -6 + 6a$$ 21. $x_3 = \frac{\det(A_3)}{\det(A)} = \frac{-6 + 6a}{6} = -1 + a$ 22. Natija: $$\boxed{\begin{cases} x_1 = -a \\ x_2 = 2 \\ x_3 = -1 + a \end{cases}}$$ 23. Bu yechim $a$ ga bog'liq, ya'ni $x_1,x_2,x_3$ ni $a$ orqali ifodaladik. 24. Agar $a$ ning qiymati ma'lum bo'lsa, $x_1,x_2,x_3$ ni aniq sonlar sifatida topish mumkin.