URGENTE!
determine o conjunto solução das equações, aplicando a regra de Sarrus
![\left[\begin{array}{ccc}-2&1&x\\1&2&4\\3&-1&-2\end{array}\right] = -7 \left[\begin{array}{ccc}-2&1&x\\1&2&4\\3&-1&-2\end{array}\right] = -7](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B1%26amp%3Bx%5C%5C1%26amp%3B2%26amp%3B4%5C%5C3%26amp%3B-1%26amp%3B-2%5Cend%7Barray%7D%5Cright%5D+%3D+-7)
Lukyo:
Determinante, certo?
Soluções para a tarefa
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Resolver a equação
![\left|\begin{array}{ccc} -2&1&x\\ 1&2&4\\ 3&-1&-2 \end{array}\right|=-7 \left|\begin{array}{ccc} -2&1&x\\ 1&2&4\\ 3&-1&-2 \end{array}\right|=-7](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D+-2%26amp%3B1%26amp%3Bx%5C%5C+1%26amp%3B2%26amp%3B4%5C%5C+3%26amp%3B-1%26amp%3B-2+%5Cend%7Barray%7D%5Cright%7C%3D-7)
usando a Regra de Sarrus.
• Primeiramente, vamos calcular o determinante do lado esquerdo da equação.
Reescrevemos as duas primeiras colunas. Depois, multiplicamos os elementos das diagonais e fazemos as somas e subtrações adequadas:
![\left|\begin{array}{ccc} -2&1&x\\ 1&2&4\\ 3&-1&-2 \end{array}\right|\begin{array}{cc} -2&1\\ 1&2\\ 3&-1 \end{array}\\\\\\ \begin{array}{rcccccr} =&-&2\cdot 2\cdot (-2)&\!\!\!+\!\!\!&1\cdot 4\cdot 3&\!\!\!+\!\!\!&x\cdot 1\cdot (-1)\\ &\!\!\!-\!\!\!&3\cdot 2\cdot x&\!\!\!-\!\!\!&(-1)\cdot 4\cdot (-2)&\!\!\!-\!\!\!&(-2)\cdot 1\cdot 1\end{array}\\\\\\ \begin{array}{lcrcrcr} =&-&(-8)&\!\!\!+\!\!\!&12&\!\!\!+\!\!\!&(-x)\\ &\!\!\!-\!\!\!&6x&\!\!\!-\!\!\!&8&\!\!\!-\!\!\!&(-2) \end{array}\\\\\\ \begin{array}{lcrcrcr} =&&8&\!\!\!+\!\!\!&12&\!\!\!-\!\!\!&x\\ &\!\!\!-\!\!\!&6x&\!\!\!-\!\!\!&8&\!\!\!+\!\!\!&2 \end{array} \left|\begin{array}{ccc} -2&1&x\\ 1&2&4\\ 3&-1&-2 \end{array}\right|\begin{array}{cc} -2&1\\ 1&2\\ 3&-1 \end{array}\\\\\\ \begin{array}{rcccccr} =&-&2\cdot 2\cdot (-2)&\!\!\!+\!\!\!&1\cdot 4\cdot 3&\!\!\!+\!\!\!&x\cdot 1\cdot (-1)\\ &\!\!\!-\!\!\!&3\cdot 2\cdot x&\!\!\!-\!\!\!&(-1)\cdot 4\cdot (-2)&\!\!\!-\!\!\!&(-2)\cdot 1\cdot 1\end{array}\\\\\\ \begin{array}{lcrcrcr} =&-&(-8)&\!\!\!+\!\!\!&12&\!\!\!+\!\!\!&(-x)\\ &\!\!\!-\!\!\!&6x&\!\!\!-\!\!\!&8&\!\!\!-\!\!\!&(-2) \end{array}\\\\\\ \begin{array}{lcrcrcr} =&&8&\!\!\!+\!\!\!&12&\!\!\!-\!\!\!&x\\ &\!\!\!-\!\!\!&6x&\!\!\!-\!\!\!&8&\!\!\!+\!\!\!&2 \end{array}](https://tex.z-dn.net/?f=%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D+-2%26amp%3B1%26amp%3Bx%5C%5C+1%26amp%3B2%26amp%3B4%5C%5C+3%26amp%3B-1%26amp%3B-2+%5Cend%7Barray%7D%5Cright%7C%5Cbegin%7Barray%7D%7Bcc%7D+-2%26amp%3B1%5C%5C+1%26amp%3B2%5C%5C+3%26amp%3B-1+%5Cend%7Barray%7D%5C%5C%5C%5C%5C%5C+%5Cbegin%7Barray%7D%7Brcccccr%7D+%3D%26amp%3B-%26amp%3B2%5Ccdot+2%5Ccdot+%28-2%29%26amp%3B%5C%21%5C%21%5C%21%2B%5C%21%5C%21%5C%21%26amp%3B1%5Ccdot+4%5Ccdot+3%26amp%3B%5C%21%5C%21%5C%21%2B%5C%21%5C%21%5C%21%26amp%3Bx%5Ccdot+1%5Ccdot+%28-1%29%5C%5C+%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B3%5Ccdot+2%5Ccdot+x%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B%28-1%29%5Ccdot+4%5Ccdot+%28-2%29%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B%28-2%29%5Ccdot+1%5Ccdot+1%5Cend%7Barray%7D%5C%5C%5C%5C%5C%5C+%5Cbegin%7Barray%7D%7Blcrcrcr%7D+%3D%26amp%3B-%26amp%3B%28-8%29%26amp%3B%5C%21%5C%21%5C%21%2B%5C%21%5C%21%5C%21%26amp%3B12%26amp%3B%5C%21%5C%21%5C%21%2B%5C%21%5C%21%5C%21%26amp%3B%28-x%29%5C%5C+%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B6x%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B8%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B%28-2%29+%5Cend%7Barray%7D%5C%5C%5C%5C%5C%5C+%5Cbegin%7Barray%7D%7Blcrcrcr%7D+%3D%26amp%3B%26amp%3B8%26amp%3B%5C%21%5C%21%5C%21%2B%5C%21%5C%21%5C%21%26amp%3B12%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3Bx%5C%5C+%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B6x%26amp%3B%5C%21%5C%21%5C%21-%5C%21%5C%21%5C%21%26amp%3B8%26amp%3B%5C%21%5C%21%5C%21%2B%5C%21%5C%21%5C%21%26amp%3B2+%5Cend%7Barray%7D)
![=20-x-6x-6\\\\ =14-7x =20-x-6x-6\\\\ =14-7x](https://tex.z-dn.net/?f=%3D20-x-6x-6%5C%5C%5C%5C+%3D14-7x)
_______
• Então para resolvermos a equação dada, devemos ter
![14-7x=-7\\\\ 14+7=7x\\\\ 7x=21\\\\ x=\dfrac{21}{7}\\\\\\ \boxed{\begin{array}{c}x=3 \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a solu\c{c}\~ao} 14-7x=-7\\\\ 14+7=7x\\\\ 7x=21\\\\ x=\dfrac{21}{7}\\\\\\ \boxed{\begin{array}{c}x=3 \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a solu\c{c}\~ao}](https://tex.z-dn.net/?f=14-7x%3D-7%5C%5C%5C%5C+14%2B7%3D7x%5C%5C%5C%5C+7x%3D21%5C%5C%5C%5C+x%3D%5Cdfrac%7B21%7D%7B7%7D%5C%5C%5C%5C%5C%5C+%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7Dx%3D3+%5Cend%7Barray%7D%7D%5Cquad%5Clongleftarrow%5Cquad%5Ctextsf%7Besta+%5C%27e+a+solu%5Cc%7Bc%7D%5C%7Eao%7D)
Conjunto solução: S = {3}.
Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7570644
Dúvidas? Comente.
Bons estudos! :-)
Tags: equação determinante regra de sarrus matriz solução resolver álgebra
usando a Regra de Sarrus.
• Primeiramente, vamos calcular o determinante do lado esquerdo da equação.
Reescrevemos as duas primeiras colunas. Depois, multiplicamos os elementos das diagonais e fazemos as somas e subtrações adequadas:
_______
• Então para resolvermos a equação dada, devemos ter
Conjunto solução: S = {3}.
Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7570644
Dúvidas? Comente.
Bons estudos! :-)
Tags: equação determinante regra de sarrus matriz solução resolver álgebra
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