Problems from Sarrus' rule: Computing third-order determinants

Invent a $3 \times 3$ matrix and compute its determinant.

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Development:

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

$d e t (B) = | \begin{matrix} 3 & 1 & - 1 \\ 0 & 1 & 2 \\ 1 & 1 & 0 \end{matrix} | = 3 \cdot 1 \cdot 0 + 0 \cdot 1 \cdot (- 1) + 1 \cdot 1 \cdot 2 - (- 1) \cdot 1 \cdot 1 - 2 \cdot 1 \cdot 3 - 0 \cdot 1 \cdot 0 =$

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

Solution:

$d e t (B) = - 3$

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Compute the following determinants.

$| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} |$ , $| \begin{matrix} 3 & 2 & 1 \\ - 2 & - 1 & 0 \\ 2 & - 5 & 0 \end{matrix} |$

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Development:

$| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \to \begin{matrix} | \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | \\ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \end{matrix} = 1 \cdot 1 \cdot 1 + 0 + 0 - 0 \cdot 1 \cdot 0 - 0 - 0 = 1$

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

Solution:

$1, 12$

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