Newton Ramirez
07/27/2024 · Junior High School
2. Calcule o valor de \( S \) : \[ S=\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 1 & 1 \\ -2 & 1 & 1\end{array}\right|+\left|\begin{array}{lll}0 & 1 & 4 \\ 0 & 2 & 5 \\ 0 & 3 & 6\end{array}\right|+\left[\left.\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 0 & -2 \\ 1 & 6 & 4\end{array} \right\rvert\,\right. \]
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
Para calcular o valor de \( S \), precisamos calcular os determinantes das matrizes fornecidas.
1. **Cálculo do primeiro determinante:**
\[
D_1 = \left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 1 & 1 \\ -2 & 1 & 1\end{array}\right|
\]
Usamos a regra de Sarrus ou a expansão por cofatores. Vamos usar a expansão pela primeira linha:
\[
D_1 = 2 \left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| - (-1) \left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right| + 1 \left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right|
\]
Calculando os determinantes 2x2:
\[
\left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| = 1 \cdot 1 - 1 \cdot 1 = 0
\]
\[
\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right| = 1 \cdot 1 - 1 \cdot (-2) = 1 + 2 = 3
\]
\[
\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right| = 3 \quad \text{(já calculado)}
\]
Substituindo:
\[
D_1 = 2 \cdot 0 + 1 \cdot 3 + 1 \cdot 3 = 0 + 3 + 3 = 6
\]
2. **Cálculo do segundo determinante:**
\[
D_2 = \left|\begin{array}{ccc}0 & 1 & 4 \\ 0 & 2 & 5 \\ 0 & 3 & 6\end{array}\right|
\]
Como a primeira coluna é composta apenas de zeros, o determinante é:
\[
D_2 = 0
\]
3. **Cálculo do terceiro determinante:**
\[
D_3 = \left|\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 0 & -2 \\ 1 & 6 & 4\end{array}\right|
\]
Usamos a expansão pela primeira linha:
\[
D_3 = -1 \left|\begin{array}{cc}0 & -2 \\ 6 & 4\end{array}\right| - 2 \left|\begin{array}{cc}5 & -2 \\ 1 & 4\end{array}\right| + 3 \left|\begin{array}{cc}5 & 0 \\ 1 & 6\end{array}\right|
\]
Calculando os determinantes 2x2:
\[
\left|\begin{array}{cc}0 & -2 \\ 6 & 4\end{array}\right| = 0 \cdot 4 - (-2) \cdot 6 = 12
\]
\[
\left|\begin{array}{cc}5 & -2 \\ 1 & 4\end{array}\right| = 5 \cdot 4 - (-2) \cdot 1 = 20 + 2 = 22
\]
\[
\left|\begin{array}{cc}5 & 0 \\ 1 & 6\end{array}\right| = 5 \cdot 6 - 0 \cdot 1 = 30
\]
Substituindo:
\[
D_3 = -1 \cdot 12 - 2 \cdot 22 + 3 \cdot 30 = -12 - 44 + 90 = 34
\]
4. **Cálculo do valor de \( S \):**
\[
S = D_1 + D_2 + D_3 = 6 + 0 + 34 = 40
\]
Portanto, o valor de \( S \) é:
\[
\boxed{40}
\]
Quick Answer
O valor de \( S \) é 40.
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