Respostas

2013-05-28T10:07:05-03:00

Olá vitor!! a solução está em anexo!! veja se entende!!!

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A melhor resposta!
2013-05-28T10:53:05-03:00

Pela Regra de Cramer, observe que está separado entre || e!! as duas colunas onde vc vai operar para encontrar as incógnitas a,b,c,d.

\ \ \ \ \ \ A\ \ \ \ \ \ \ .A^-^1\ \ \ \ \ \ \ =\ \ \ \ \ I\\\left[\begin{array}{ccc}3&\ 1\\2&-1\end{array}\right].\left[\begin{array}{ccc}a&c\\b&d\end{array}\right]=\left[\begin{array}{ccc}|1|&!0!\\|0|&!1!\end{array}\right]\\\\A^-^1a:\left[\begin{array}{ccc}|1|&1\\|0|&-1\end{array}\right]=>det(A^-^1a)=-1\\\\A^-^1b:\left[\begin{array}{ccc}3&|1|\\2&|0|\end{array}\right]=>det(A^-^1b)=-2\\\\A^-^1c:\left[\begin{array}{ccc}!0!&1\\!1!&-1\end{array}\right]=>det(A^-^1c)=-1\\\\

A^-^1d:\left[\begin{array}{ccc}3&!0!\\2&!1!\end{array}\right]=>det(A^-^1d)=3\\\\ A:\left[\begin{array}{ccc}3&1\\2&-1\end{array}\right]=>det(A)=-5\\\\

A^-^1d:\left[\begin{array}{ccc}3&!0!\\2&!1!\end{array}\right]=>det(A^-^1d)=3\\\\ A:\left[\begin{array}{ccc}3&1\\2&-1\end{array}\right]=>det(A)=-5\\\\A^-^1:\left[\begin{array}{ccc}\frac{det(A^-^1a)}{det(A)}&\frac{det(A^-^1c)}{det(A)}\\\frac{det(A^-^1b)}{Det(A)}&\frac{det(A^-^1d)}{det(A)}\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{-5}&\frac{-1}{-5}\\\frac{-2}{-5}&\frac{3}{-5}\end{array}\right]=>\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{-3}{5}\end{array}\right]

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