1. Muammo: 4x4 matritsaning determinantini hisoblash kerak. Matritsa: $$\begin{pmatrix} 3 & -3 & -2 & -5 \\ 2 & -5 & 4 & 6 \\ 5 & 5 & 8 & 7 \\ 4 & 4 & 5 & 6 \end{pmatrix}$$ 2. Laplas teoremasi yordamida determinantni hisoblash: Laplas teoremasi bo'yicha, determinantni bir qator yoki ustun bo'yicha kengaytiramiz. Masalan, birinchi qator bo'yicha: $$\det(A) = \sum_{j=1}^4 (-1)^{1+j} a_{1j} M_{1j}$$ bu yerda $a_{1j}$ - birinchi qator elementlari, $M_{1j}$ - shu elementga mos kichik determinant (minor). 3. Har bir $M_{1j}$ ni hisoblaymiz: - $M_{11}$ - 1-qator va 1-ustun chiqarilgan matritsaning determinantini hisoblaymiz: $$\begin{vmatrix} -5 & 4 & 6 \\ 5 & 8 & 7 \\ 4 & 5 & 6 \end{vmatrix}$$ - $M_{12}$ - 1-qator va 2-ustun chiqarilgan matritsa: $$\begin{vmatrix} 2 & 4 & 6 \\ 5 & 8 & 7 \\ 4 & 5 & 6 \end{vmatrix}$$ - $M_{13}$ - 1-qator va 3-ustun chiqarilgan matritsa: $$\begin{vmatrix} 2 & -5 & 6 \\ 5 & 5 & 7 \\ 4 & 4 & 6 \end{vmatrix}$$ - $M_{14}$ - 1-qator va 4-ustun chiqarilgan matritsa: $$\begin{vmatrix} 2 & -5 & 4 \\ 5 & 5 & 8 \\ 4 & 4 & 5 \end{vmatrix}$$ 4. Har bir 3x3 determinantni Sarrus qoidasi yoki Laplas teoremasi bilan hisoblaymiz. Misol uchun, $M_{11}$: $$\det(M_{11}) = (-5)(8 \cdot 6 - 7 \cdot 5) - 4(5 \cdot 6 - 7 \cdot 4) + 6(5 \cdot 5 - 8 \cdot 4)$$ $$= -5(48 - 35) - 4(30 - 28) + 6(25 - 32)$$ $$= -5(13) - 4(2) + 6(-7) = -65 - 8 - 42 = -115$$ 5. Shu tarzda qolgan minorlarni ham hisoblaymiz: $M_{12}$: $$2(8 \cdot 6 - 7 \cdot 5) - 4(5 \cdot 6 - 7 \cdot 4) + 6(5 \cdot 5 - 8 \cdot 4)$$ $$= 2(48 - 35) - 4(30 - 28) + 6(25 - 32) = 2(13) - 4(2) + 6(-7) = 26 - 8 - 42 = -24$$ $M_{13}$: $$2(5 \cdot 6 - 7 \cdot 4) - (-5)(5 \cdot 6 - 7 \cdot 4) + 6(5 \cdot 4 - 5 \cdot 4)$$ $$= 2(30 - 28) + 5(30 - 28) + 6(20 - 20) = 2(2) + 5(2) + 6(0) = 4 + 10 + 0 = 14$$ $M_{14}$: $$2(5 \cdot 5 - 8 \cdot 4) - (-5)(5 \cdot 5 - 8 \cdot 4) + 4(5 \cdot 4 - 5 \cdot 4)$$ $$= 2(25 - 32) + 5(25 - 32) + 4(20 - 20) = 2(-7) + 5(-7) + 4(0) = -14 - 35 + 0 = -49$$ 6. Endi Laplas formulaga qo'yamiz: $$\det(A) = 3 \cdot (-1)^{1+1} \cdot (-115) + (-3) \cdot (-1)^{1+2} \cdot (-24) + (-2) \cdot (-1)^{1+3} \cdot 14 + (-5) \cdot (-1)^{1+4} \cdot (-49)$$ $$= 3 \cdot 1 \cdot (-115) + (-3) \cdot (-1) \cdot (-24) + (-2) \cdot 1 \cdot 14 + (-5) \cdot (-1) \cdot (-49)$$ $$= -345 - 72 - 28 - 245 = -690$$ 7. Elementar almashtirish usuli bilan hisoblash: Matritsani qatorlar bilan elementar almashtirishlar yordamida yuqori uchburchak ko'rinishga keltirib, determinantni hisoblash mumkin. Determinant qator almashtirishda ishorasi o'zgaradi, qatorni ko'paytirishda esa determinant ko'paytiriladi. Bu usulda ham natija $-690$ chiqadi. Javob: $$\boxed{-690}$$