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1. Determine o conjunto solução dos seguintes sistemas linea...
- Determine o conjunto solução dos seguintes sistemas lineares reais, usando os algoritmos
de eliminação e eliminação total:
2x+y+z &= 0 \\
x+3z+4y &= 1 \\
4(y+x) &= 2
\end{cases}$$
$$\begin{cases}
x+y-w &= 1 \\
x-y+z &= 1 \\
x-y+w &= 1
\end{cases}$$
$$\begin{cases}
x_1-x_2-x_3+x_4 &= 1 \\
x_1+x_2-x_3-x_4 &= 1 \\
x_1-x_2+x_3-x_4 &= -1 \\
-x_1+x_2-x_3+x_4 &= 0
\end{cases}$$
$$\begin{cases}
x_1+x_4 &= 1+x_2+x_3 \\
x_1+x_2 &= 1+x_4+x_4 \\
x_1+x_3 &= 1+x_2+x_4 \\
x_2+x_4 &= 1+x_1+x_3
\end{cases}$$
$$\begin{cases}
x+y &= 1 \\
y+z &= 1 \\
z+w &= 1 \\
w+x &= 1 \\
x+y+z+w &= 2
\end{cases}$$
$$\begin{cases}
x+y &= 2 \\
y+z &= 0 \\
z+w &= 2 \\
w+x &= 0 \\
x+y+z+w &= 4
\end{cases}$$
$$\begin{cases}
x_3+x_4 &= -1 \\
x_1+x_2 &= 1
\end{cases}$$
$$\begin{cases}
2y+x-2z &= -1 \\
2-2x-y+z &= 1 \\
2z+x &= 3
\end{cases}$$
$$\begin{cases}
2x + y &= 2z+y \\
3x-2y+4 &=1-z \\
-x+2y &= 2
\end{cases}$$
$$\begin{cases}
x_1+x_2+x_3 &= 0 \\
-x_1+x_2-x_3 &= 0 \\
x_1+x_2+x_3 &= -1
\end{cases}$$
$$\begin{cases}
x+7y+z &=3-2z \\
3x+y+2z&=1-3z \\
2x-2y &= 2-3z
\end{cases}$$
$$\begin{cases}
2x + y &= 0 \\
2y+x+z &= 0 \\
2z+y &= 0
\end{cases}$$
$$\begin{cases}
x_3+x_4 &=-1 \\
x_1+x_2 &= 1 \\
x_2+x_3 &= 2 \\
x_4+x_1 &= 0
\end{cases}$$
$$\begin{cases}
2x + y &= y \\
y+x+z &= y \\
2z+y &= y
\end{cases}$$
$$\begin{cases}
-2x+y+z&=0 \\
3y+4z &=1 \\
4x+2y &=2 \\
w-x &=2
\end{cases}$$
$$\begin{cases}
2x+y+z &= 0 \\
x+3y+4z &= 1
\end{cases}$$
$$\begin{cases}
x+2y+3w+4z &= 0
\end{cases}$$
1. Exemplo:
(a) Matriz das variáveis: $\begin{bmatrix} x \\ y \\ z \end{bmatrix} \in \mathbb{R}^{3x1}$
1
Organizando o sistema na forma padrão temos $\begin{cases}
2x+y+z &= 0 \\
x+4y+3z &= 1 \text{ com } \\
4x+4y+0z&=2
\end{cases}$
matriz associada $\begin{bmatrix}
2 & 1 & 1 & 0 \\
1 & 4 & 3 & 1 \\
4 & 4 & 0 & 2
\end{bmatrix}$
Escalonando a matriz do sistema:
$\begin{bmatrix}
2 & 1 & 1 & 0 \\
1 & 4 & 3 & 1 \\
4 & 4 & 0 & 2
\end{bmatrix} \leftrightarrow \begin{bmatrix}
1 & 4 & 3 & 1 \\
2 & 1 & 1 & 0 \\
4 & 4 & 0 & 2
\end{bmatrix} \xrightarrow[\,-4]{-2} \begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & -12 & -12 & -2
\end{bmatrix} \xrightarrow{\frac{12}{7}} \begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & 0 & \frac{24}{7} & \frac{10}{7}
\end{bmatrix} | \begin{matrix} \\\ -\frac{12}{7} \\\ -\frac{7}{24} \end{matrix} \leftrightarrow \begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix}$
Logo o sistema tem solução e a solução é única.
Escalonando totalmente a matriz do sistema a partir da matriz escalonada:
$\begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix} \xleftarrow[\,5]{} \begin{bmatrix}
1 & 3 & 0 & \frac{27}{12} \\\ 0 & -7 & 0 & -\frac{25}{12} \\\ 0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix} \xrightarrow{-\frac{3}{7}} \begin{bmatrix}
1 & 0 & 0 & \frac{114}{84} \\\ 0 & -7 & 0 & -\frac{25}{12} \\\ 0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix} \xrightarrow{-\frac{1}{7}} \begin{bmatrix}
1 & 0 & 0 & \frac{114}{84} \\\ 0 & 1 & 0 & \frac{25}{84} \\\ 0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix}$
- Determine o conjunto solução dos seguintes sistemas lineares reais, usando os algoritmos de eliminação e eliminação total:
Responde essas questões de acordo com o exemplo abaixo delas
![1. Determine o conjunto solução dos seguintes sistemas lineares reais, usando os algoritmos
de eliminação e eliminação total:
$$\begin{cases}
2x+y+z &= 0 \\
x+3z+4y &= 1 \\
4(y+x) &= 2
\end{cases}$$
$$\begin{cases}
x+y-w &= 1 \\
x-y+z &= 1 \\
x-y+w &= 1
\end{cases}$$
$$\begin{cases}
x_1-x_2-x_3+x_4 &= 1 \\
x_1+x_2-x_3-x_4 &= 1 \\
x_1-x_2+x_3-x_4 &= -1 \\
-x_1+x_2-x_3+x_4 &= 0
\end{cases}$$
$$\begin{cases}
x_1+x_4 &= 1+x_2+x_3 \\
x_1+x_2 &= 1+x_4+x_4 \\
x_1+x_3 &= 1+x_2+x_4 \\
x_2+x_4 &= 1+x_1+x_3
\end{cases}$$
$$\begin{cases}
x+y &= 1 \\
y+z &= 1 \\
z+w &= 1 \\
w+x &= 1 \\
x+y+z+w &= 2
\end{cases}$$
$$\begin{cases}
x+y &= 2 \\
y+z &= 0 \\
z+w &= 2 \\
w+x &= 0 \\
x+y+z+w &= 4
\end{cases}$$
$$\begin{cases}
x_3+x_4 &= -1 \\
x_1+x_2 &= 1
\end{cases}$$
$$\begin{cases}
2y+x-2z &= -1 \\
2-2x-y+z &= 1 \\
2z+x &= 3
\end{cases}$$
$$\begin{cases}
2x + y &= 2z+y \\
3x-2y+4 &=1-z \\
-x+2y &= 2
\end{cases}$$
$$\begin{cases}
x_1+x_2+x_3 &= 0 \\
-x_1+x_2-x_3 &= 0 \\
x_1+x_2+x_3 &= -1
\end{cases}$$
$$\begin{cases}
x+7y+z &=3-2z \\
3x+y+2z&=1-3z \\
2x-2y &= 2-3z
\end{cases}$$
$$\begin{cases}
2x + y &= 0 \\
2y+x+z &= 0 \\
2z+y &= 0
\end{cases}$$
$$\begin{cases}
x_3+x_4 &=-1 \\
x_1+x_2 &= 1 \\
x_2+x_3 &= 2 \\
x_4+x_1 &= 0
\end{cases}$$
$$\begin{cases}
2x + y &= y \\
y+x+z &= y \\
2z+y &= y
\end{cases}$$
$$\begin{cases}
-2x+y+z&=0 \\
3y+4z &=1 \\
4x+2y &=2 \\
w-x &=2
\end{cases}$$
$$\begin{cases}
2x+y+z &= 0 \\
x+3y+4z &= 1
\end{cases}$$
$$\begin{cases}
x+2y+3w+4z &= 0
\end{cases}$$
1. Exemplo:
(a) Matriz das variáveis: $\begin{bmatrix} x \\ y \\ z \end{bmatrix} \in \mathbb{R}^{3x1}$
1
Organizando o sistema na forma padrão temos $\begin{cases}
2x+y+z &= 0 \\
x+4y+3z &= 1 \text{ com } \\
4x+4y+0z&=2
\end{cases}$
matriz associada $\begin{bmatrix}
2 & 1 & 1 & 0 \\
1 & 4 & 3 & 1 \\
4 & 4 & 0 & 2
\end{bmatrix}$
Escalonando a matriz do sistema:
$\begin{bmatrix}
2 & 1 & 1 & 0 \\
1 & 4 & 3 & 1 \\
4 & 4 & 0 & 2
\end{bmatrix} \leftrightarrow \begin{bmatrix}
1 & 4 & 3 & 1 \\
2 & 1 & 1 & 0 \\
4 & 4 & 0 & 2
\end{bmatrix} \xrightarrow[\,-4]{-2} \begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & -12 & -12 & -2
\end{bmatrix} \xrightarrow{\frac{12}{7}} \begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & 0 & \frac{24}{7} & \frac{10}{7}
\end{bmatrix} | \begin{matrix} \\\ -\frac{12}{7} \\\ -\frac{7}{24} \end{matrix} \leftrightarrow \begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix}$
Logo o sistema tem solução e a solução é única.
Escalonando totalmente a matriz do sistema a partir da matriz escalonada:
$\begin{bmatrix}
1 & 4 & 3 & 1 \\
0 & -7 & -5 & -2 \\
0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix} \xleftarrow[\,5]{} \begin{bmatrix}
1 & 3 & 0 & \frac{27}{12} \\\ 0 & -7 & 0 & -\frac{25}{12} \\\ 0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix} \xrightarrow{-\frac{3}{7}} \begin{bmatrix}
1 & 0 & 0 & \frac{114}{84} \\\ 0 & -7 & 0 & -\frac{25}{12} \\\ 0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix} \xrightarrow{-\frac{1}{7}} \begin{bmatrix}
1 & 0 & 0 & \frac{114}{84} \\\ 0 & 1 & 0 & \frac{25}{84} \\\ 0 & 0 & 1 & -\frac{5}{12}
\end{bmatrix}$](https://prod-meuguru-guruia-assets.s3.sa-east-1.amazonaws.com/YNlu2jY1uuCXmG6-orcQIt9Im?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Content-Sha256=UNSIGNED-PAYLOAD&X-Amz-Credential=ASIAQXTC4KK66XVLYIVD%2F20250716%2Fsa-east-1%2Fs3%2Faws4_request&X-Amz-Date=20250716T063010Z&X-Amz-Expires=900&X-Amz-Security-Token=IQoJb3JpZ2luX2VjED0aCXNhLWVhc3QtMSJIMEYCIQDtrXBVmVAEcsuvcOv8INybcDUdunEXqh2aIxh2Uwl2rgIhAPogQ5f%2Fuz7xn7zNRF8KdDQK%2FU78tLFo30RYRw9z3MNnKuUDCFYQAhoMMDUwNjczMjQyODEzIgzj4u%2Fiwfl1CUU5F8wqwgPY89oR40GOsE4dJNRb9de971RLfjrbt4hxeOhTFwq9APBDDezK99mfKLDV2OBBFz25QdHxV9zLMfBYwTUvja0%2FSAEAmFAyptMt7upduzXm4m4OJUSSsQSU%2BZYeiaAzsrAdHFS4Ggv3omjRuNPOUWXnrgmkNpW53tkZsTH%2F9JsfVPdYt6YFPHDZYIhdhywvXgmebpCNJvy4RgdGZM6B6oDWd9tt3BGu9tS6fvKR4ZB2ER0yUcrj0vGZsGNklkLyEmXcreXi8g%2Ff%2FiE0HmJOMsvllBn5cbdcmDl%2B5mJrUHhvXD5sJkj0qJVwcv8KnyFHD9HTz%2BhtfDwQEnKIplgd%2BfhiM5Aj3BQhUT%2Fib%2Bhcab46nqisGCXAibwxYOfBmba6q2rGyD440Jgil7CHfbmBdRr8Jo%2BcNy9xye53MEuZXLaT3vN0boBAc6s%2FaXxRz7tBFDyC1caqsY3pAcDre1odXDQ4vU7bAaGwMl78GOGy9n%2Fhda10I%2BHh%2BxjRlv0Wxzh8cGK%2FcCYQQF3FSKpCn2TKMFg7mEQceXc9t7sT9wzC03oSgfJdBMXd4gzzbCjYWfmNrIxrQjPaCQ79DN94mJ5kC8IvLiQw09zcwwY6pAEpICbJcKzYjF59nYYZgNIeU9%2Ff030gq7%2FiF768%2BnJZnzFAQNYlp7Eh%2BktTlgWgeEZyklgFli%2FN5TVo94xguu%2FJkUFQNqcxGw6JhTKb9WHK2H5YjDY%2FPGBD%2FTYVhTC5kBhMQMQocdepew%2FmSWdaXG3HWM9z24zpYTTE6Ex9%2Fp7Jh3CH3uK5rPd8WKQ3PzDB5KSpj153YdMIrcUwePO%2FA2BjKDNh8w%3D%3D&X-Amz-Signature=7a4ca4c8f7e5cdac2a75e5ee37a7417bcaf2e3c32d7cca36dd099af127ece7cb&X-Amz-SignedHeaders=host&x-id=GetObject)
