{lll}{1}&{5}&{1}{2}&{3}&{0}{4}&{2}&{1} eching

$$\left( \begin{array} { l l l } { 1 } & { 5 } & { 1 } \\ { 2 } & { 3 } & { 0 } \\ { 4 } & { 2 } & { 1 } \end{array} \right)$$

Determinantni hisoblash

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Sarrus qoidasidan foydalanish qadamlari

Diagonal usulidan foydalanib matritsaning aniqlovchisini topish.

$$det(\left(\begin{matrix}1&5&1\\2&3&0\\4&2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&5&1&1&5\\2&3&0&2&3\\4&2&1&4&2\end{matrix}\right)$$

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

$$3+2\times 2=7$$

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

$$4\times 3+2\times 5=22$$

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

$$7-22$$

7 dan 22 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).

$$det(\left(\begin{matrix}1&5&1\\2&3&0\\4&2&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}3&0\\2&1\end{matrix}\right))-5det(\left(\begin{matrix}2&0\\4&1\end{matrix}\right))+det(\left(\begin{matrix}2&3\\4&2\end{matrix}\right))$$

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

$$3-5\times 2+2\times 2-4\times 3$$

Qisqartirish.

$$3-5\times 2-8$$

Yakuniy natija olish uchun shartlarni qo'shish.

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Baholash

$$\left(\begin{matrix}1&5&1\\2&3&0\\4&2&1\end{matrix}\right)$$