@import url(https://fonts.googleapis.com/css?family=Source+Sans+Pro:300,400,600,700); Questão 59 \u2013 Para que o determinante da matriz \ufeff\ufeff\ufeff(1amp;\u22121amp;11amp;0amp;b1amp;2amp;1)\ufeff\ufeff\ufeff\begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & b\\ 1& 2&1\end{pmatrix} \ufeff \ufeff\ufeff\u239d\u239b\u200b111\u200bamp;\u22121amp;0amp;2\u200bamp;1amp;bamp;1\u200b\u23a0\u239e\u200b\ufeff\ufeff seja 3, o valor de b deve ser igual a: a) 2 b) 0 c) -1 d) -2 RESOLUÇÃO Fala galera, para resolvermos essa questão precisamos lembrar como faz determinante de uma matriz."Nas matrizes quadradas de ordem 3x3 esses cálculos podem ser efetuados repetindo-se a 1ª e a 2ª coluna, aplicando em seguida a regra de Sarrus."\ufeff\u22231amp;\u22121amp;11amp;0amp;b1amp;2amp;1\u2223\left| \begin{array}{c c c} 1&-1&1\\ 1& 0& b\\ 1&2&1\end{array}\right|\u2223\u2223\u2223\u2223\u2223\u2223\u200b111\u200bamp;\u22121amp;0amp;2\u200bamp;1amp;bamp;1\u200b\u2223\u2223\u2223\u2223\u2223\u2223\u200b\ufeff \ufeff\u22231amp;\u221211amp;01amp;2\u2223=3\left| \begin{array}{c c } 1&-1\\ 1& 0\\ 1 & 2 \end{array}\right| = 3\u2223\u2223\u2223\u2223\u2223\u2223\u200b111\u200bamp;\u22121amp;0amp;2\u200b\u2223\u2223\u2223\u2223\u2223\u2223\u200b=3\ufeff \ufeff1\u22c50\u22c51+\u22121\u22c5b\u22c51+1\u22c51\u22c52\u22121\u22c50\u22c51\u22122\u22c5b\u22c51\u22121\u22c51\u22c5(\u22121)1 \cdot 0 \cdot 1 + -1 \cdot b \cdot 1 + 1 \cdot 1 \cdot 2 - 1\cdot 0 \cdot 1 - 2 \cdot b \cdot 1 - 1 \cdot 1 \cdot (-1)1\u22c50\u22c51+\u22121\u22c5b\u22c51+1\u22c51\u22c52\u22121\u22c50\u22c51\u22122\u22c5b\u22c51\u22121\u22c51\u22c5(\u22121)\ufeff=3 \ufeff\u2212b+2\u22122b+1=3-b+2-2b+1=3\u2212b+2\u22122b+1=3\ufeff \ufeff\u22123b+3=3-3b+3=3\u22123b+3=3\ufeff \ufeffb=0b=0b=0\ufeff GABARITO LETRA BQualquer dúvida, meu instagram @carol.1111
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