{lll}{1}&{2}&{4}{1}&{4}&{1}{6}&{8}&{5} eching

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 4 } \\ { 1 } & { 4 } & { 1 } \\ { 6 } & { 8 } & { 5 } \end{array} \right|$$

Baholash

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Sarrus qoidasidan foydalanish qadamlari

Diagonal usulidan foydalanib matritsaning aniqlovchisini topish.

$$det(\left(\begin{matrix}1&2&4\\1&4&1\\6&8&5\end{matrix}\right))$$

Birinchi ikki ustunni to'rtinchi va beshinchi ustun sifatida qaytarish uchun asl matritsani kengaytirish.

$$\left(\begin{matrix}1&2&4&1&2\\1&4&1&1&4\\6&8&5&6&8\end{matrix}\right)$$

Pastki chap kiritmadan boshlab diagonal kamayish tarafga ko'paytirish ha koʻpaytmalarni qo'shish.

$$4\times 5+2\times 6+4\times 8=64$$

Pastki chap kiritmadan boshlab diagonal ko'payish tarafga koʻpaytirish va koʻpaytmalarni qoʻshish.

$$6\times 4\times 4+8+5\times 2=114$$

Kamayib boruvchi diagonal koʻpaytmalarning yig'indisidan ko'payib boruvchi diagonal koʻpaytmalar yig'indisini ayirish.

$$64-114$$

64 dan 114 ni ayirish.

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Ikkilamchilar bilan kengayishdan foydalanish qadamlari

Kichik a'zolarni kengaytirish usulidan foydalanib matritsaning aniqlovchisini topish (ko'paytiruvchilarni kengaytirish deb ham nomlanadi).

$$det(\left(\begin{matrix}1&2&4\\1&4&1\\6&8&5\end{matrix}\right))$$

Kichik a’zolarga kengaytirish uchun birinchi satrning har bir elementini o‘zining kichik a’zosiga ko‘paytiring, qaysiki ana shu elementga ega bo‘lgan satr va ustunni yo‘q qilish orqali yaratilgan 2\times 2 matritsasining maxraji hisoblanadi, so‘ngra elementning joylashuv belgisiga ko‘paytiring.

$$det(\left(\begin{matrix}4&1\\8&5\end{matrix}\right))-2det(\left(\begin{matrix}1&1\\6&5\end{matrix}\right))+4det(\left(\begin{matrix}1&4\\6&8\end{matrix}\right))$$

\left(\begin{matrix}a&b\\c&d\end{matrix}\right) 2\times 2 matritsasi uchun, determinant ad-bc.

$$4\times 5-8-2\left(5-6\right)+4\left(8-6\times 4\right)$$

Qisqartirish.

$$12-2\left(-1\right)+4\left(-16\right)$$

Yakuniy natija olish uchun shartlarni qo'shish.

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Omil

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