Black Burgess
08/01/2024 · Primary School
2. Calcule o valor de S: \[ S=\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 1 & 1 \\ -2 & 1 & 1\end{array}\right|+\left|\begin{array}{lll}0 & 1 & 4 \\ 0 & 2 & 5 \\ 0 & 3 & 6\end{array}\right|+\left|\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 0 & -2 \\ 1 & 6 & 4\end{array}\right| \]
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Para calcular o valor de \( S \), precisamos calcular cada determinante individual e somar os resultados. Vamos calcular cada um deles um por um.
### Determinante da primeira matriz:
\[ \left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 1 & 1 \\ -2 & 1 & 1\end{array}\right| \]
Usando a fórmula do determinante de uma matriz 3x3:
\[ \text{Determinante} = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]
\[ = 2(1 \cdot 1 - 1 \cdot 1) - (-1)(1 \cdot 1 - 1 \cdot (-2)) + 1(1 \cdot 1 - 1 \cdot 1) \]
\[ = 2(1 - 1) - (-1)(1 + 2) + 1(1 - 1) \]
\[ = 2 \cdot 0 - (-1) \cdot 3 + 1 \cdot 0 \]
\[ = 0 + 3 + 0 \]
\[ = 3 \]
### Determinante da segunda matriz:
\[ \left|\begin{array}{lll}0 & 1 & 4 \\ 0 & 2 & 5 \\ 0 & 3 & 6\end{array}\right| \]
Para uma matriz 3x3 onde a primeira linha é toda zero, o determinante é zero. Portanto:
\[ \left|\begin{array}{lll}0 & 1 & 4 \\ 0 & 2 & 5 \\ 0 & 3 & 6\end{array}\right| = 0 \]
### Determinante da terceira matriz:
\[ \left|\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 0 & -2 \\ 1 & 6 & 4\end{array}\right| \]
Usando a fórmula do determinante de uma matriz 3x3:
\[ \text{Determinante} = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]
\[ = (-1)(0 \cdot 4 - (-2) \cdot 6) - 2(5 \cdot 4 - (-2) \cdot 1) + 3(5 \cdot 6 - 0 \cdot 1) \]
\[ = (-1)(0 + 12) - 2(20 + 2) + 3(30) \]
\[ = (-1)(12) - 2(22) + 3(30) \]
\[ = -12 - 44 + 90 \]
\[ = 34 \]
Agora, somamos os determinantes:
\[ S = 3 + 0 + 34 \]
\[ S = 37 \]
Portanto, o valor de \( S \) é 37.
Quick Answer
\[ S = 37 \]
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