Question

Resolva o sistema linear utilizando o método de eliminação de Gauss–Jordan
!
x + 3y − z = −2
2x + y + z = −3
3x − y + z= 2
.

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\begin{cases}\mathsf{x + 3y - z = -2}\\\mathsf{2x + y + z = -3}\\\mathsf{3x - y + z = 2}\end{cases}$

$\begin{bmatrix}\cancel1&\cancel3&\cancel-1&\cancel-2\\\cancel2&\cancel1&\cancel1&\cancel -3\\\cancel3&\cancel-1&\cancel1&\cancel2\end{bmatrix}$

$\mathsf{(-1)L_2 + 2L_1}$

$\mathsf{(-1)L_3 + 3L_1}$

$\begin{bmatrix}\cancel1&\cancel3&\cancel-1&\cancel-2\\\cancel0&\cancel5&\cancel-3&\cancel 1\\\cancel0&\cancel10&\cancel-4&\cancel8\end{bmatrix}$

$\mathsf{(-5)L_1 + 3L_2}$

$\mathsf{(-1)L_3 + 2L_2}$

$\begin{bmatrix}\cancel-5&\cancel0&\cancel-4&\cancel7\\\cancel0&\cancel5&\cancel-3&\cancel -1\\\cancel0&\cancel0&\cancel-2&\cancel6\end{bmatrix}$

$\mathsf{(-1)L_1 + 2L_3}$

$\mathsf{(-2)L_2 + 3L_3}$

$\begin{bmatrix}\cancel5&\cancel0&\cancel0&\cancel5\\\cancel0&\cancel-10&\cancel0&\cancel 20\\\cancel0&\cancel0&\cancel-2&\cancel6\end{bmatrix}$

$\mathsf{\left(\dfrac{1}{5}\right)L_1}$

$\mathsf{\left(-\dfrac{1}{10}\right)L_2}$

$\mathsf{\left(-\dfrac{1}{2}\right)L_3}$

$\begin{bmatrix}\cancel1&\cancel0&\cancel0&\cancel1\\\cancel0&\cancel1&\cancel0&\cancel -2\\\cancel0&\cancel0&\cancel1&\cancel-3\end{bmatrix}$

$\boxed{\boxed{\mathsf{x = 1}}}$

$\boxed{\boxed{\mathsf{y = -2}}}$

$\boxed{\boxed{\mathsf{z = -3}}}$