Matemática, perguntado por marialauane74, 1 ano atrás

x+y=1 -2x+3y -3z = 2 x + z = 1 por escalonamento

Soluções para a tarefa

Respondido por GeBEfte
0

\left\{\begin{array}{ccccc}x&+~~y&&=&1\\-2x&~+~~3y&-3z&=&2\\x&&+~~z&=&1\end{array}\right\\\\\\Montando~a~Matriz\\\\\\\left[\begin{array}{ccccc}1&1&0&|&1\\-2&3&-3&|&2\\1&0&1&|&1\end{array}\right]\\\\\\L2\leftarrow L2+2L1\\\\L3\leftarrow L3-L1\\\\\\\left[\begin{array}{ccccc}1&1&0&|&1\\0&5&-3&|&4\\0&-1&1&|&0\end{array}\right]\\\\\\L3\leftarrow 5L3+L2\\\\\\\left[\begin{array}{ccccc}1&1&0&|&1\\0&5&-3&|&4\\0&0&2&|&4\end{array}\right]

Com a matriz escalonada, podemos remontar o sistema e determinar as variáveis:

\left\{\begin{array}{ccccc}x&+~~y&&=&1\\&~+~~5y&-3z&=&4\\&&2z&=&4\end{array}\right\\\\\\Pela~3^a~equacao:\\\\2z~=~4\\\\\\z~=~\frac{4}{2}\\\\\\\boxed{z~=~2}\\\\\\Pela~2^a~equacao:\\\\5y-3z~=~4\\\\\\5y~-~3~.~(2)~=~4\\\\\\5y-6~=~4\\\\\\5y~=~10\\\\\\y~=~\frac{10}{5}\\\\\\\boxed{y~=~2}\\\\\\Pela~3^a~equacao:\\\\x+y~=~1\\\\\\x+2~=~1\\\\\\\boxed{x~=~-1}

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