Problems from Sarrus' rule: Computing third-order determinants

Development:

$$B=\left(\begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \\ 1 & 1 & 0 \end{matrix} \right)$$

$$det(B)=\left|\begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \\ 1 & 1 & 0 \end{matrix} \right|=3\cdot1\cdot0+0\cdot1\cdot(-1)+1\cdot1\cdot2-(-1)\cdot1\cdot1-2\cdot1\cdot3-0\cdot1\cdot0=$$

$$=0+0+2+1-6-0=-3$$

Solution:

$$det(B)=-3$$

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Development:

$$\left|\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right| \rightarrow \begin{matrix} \left|\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix}\right| \\ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}\end{matrix}=1\cdot1\cdot1+0+0-0\cdot1\cdot0-0-0=1$$

$$\begin{matrix} \left|\begin{matrix} 3 & 2 & 1\\ -2 & -1 & 0\\ 2 & -5 & 0 \end{matrix}\right| \\ \begin{matrix} 3 & 2 & 1 \\ -2 & -1 & 0 \end{matrix}\end{matrix}=0+(-2)\cdot(-5)\cdot1+0-1\cdot(-1)\cdot2-0-0=12$$

Solution:

$$1, 12$$

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