Question

resolva os sistemas lineares abaixo:

a) x - 2y - 2z = -1
x - y + z = -2
2x + y + 3z= 1


b) x + 2y - z= 2
2x - y + 3z= 9
3x + 3y - 2z= 3

c) 2x + 3y + z= 11
x + y + z= 6
5x + 2y + 3z= 18​

Answer 1

Para resolver os sistemas lineares de 3 incógnitas, podemos utilizar matrizes. Transformamos as equações em uma matriz e calculamos seu determinante (D) utilizando a Regra de Sarrus.

Regra de Sarrus: soma dos produtos da diagonal principal subtraída da soma dos produtos da diagonal secundária.

a) $\left[\begin{array}{ccc}1&-2&-2\\1&-1&1\\2&1&3\end{array}\right]$

Calculamos o determinante:

$\left|\begin{array}{ccc}1&-2&-2\\1&-1&1\\2&1&3\end{array}\right| \begin{array}{ccc}1&-2&\\1&-1&\\2&1&\end{array}$

D = -3 + (-4) + (-2) - (-6) - 1 - (+4) = - 3 - 4 - 2 + 6 - 1 - 4 = -8

Calculamos agora $D_x$, $D_y$ e $D_z$, substituindo os termos independentes da equação nas respectivas colunas. Neste caso, os termos independentes são -1, -2 e 1 (o que está à direita da equação).

$D_x=\left|\begin{array}{ccc}-1&-2&1\\-2&-1&1\\1&1&3\end{array}\right| \begin{array}{ccc}-1&-2&\\-2&-1&\\1&1&\end{array}$

$D_x$ = 3 - 2 + 4 - 12 - (-1) - 2 = 3 - 2 + 4 - 12 + 1 - 2 = -8

$D_y=\left|\begin{array}{ccc}1&-1&-2\\1&-2&1\\2&1&3\end{array}\right| \begin{array}{ccc}1&-1&\\1&-2&\\2&1&\end{array}$

$D_y$ = -6 - 2 - 2 - (-3) - 1 - 8 = -6 - 2 - 2 + 3 - 1 - 8 = -16

$D_z=\left|\begin{array}{ccc}1&-2&-1\\1&-1&-2\\2&1&1\end{array}\right| \begin{array}{ccc}1&-2&\\1&-1&\\2&1&\end{array}$

$D_z$ = -1 + 8 - 1 - (-2) - (-2) - 2 = -1 + 8 - 1 + 2 + 2 - 2 = 8

Para achar os valores de x, y e z, fazemos:

$\left\{ \begin{array}{c}x=\frac{D_x}{D}=\frac{-8}{-8} = 1 \\\\y =\frac{D_y}{D}=\frac{-16}{-8}=2\\\\z=\frac{D_z}{D}=\frac{8}{-8}=-1\\\end{array}\right.$

b) $\left[\begin{array}{ccc}1&2&-1\\2&-1&3\\3&3&-2\end{array}\right]$

$D=\left|\begin{array}{ccc}1&2&-1\\2&-1&3\\3&3&-2\end{array}\right| \begin{array}{ccc}1&2&\\2&-1&\\3&3&\end{array}$

D = 10

Os termos independentes são 2, 9 e 3.

$D_x=\left|\begin{array}{ccc}2&2&-1\\9&-1&3\\3&3&-2\end{array}\right| \begin{array}{ccc}2&2&\\9&-1&\\3&3&\end{array}$

$D_x$ = 10

$D_y=\left|\begin{array}{ccc}1&2&-1\\2&9&3\\3&3&-2\end{array}\right| \begin{array}{ccc}1&2&\\2&9&\\3&3&\end{array}$

$D_y$ = 20

$D_z=\left|\begin{array}{ccc}1&2&2\\2&-1&9\\3&3&3\end{array}\right| \begin{array}{ccc}1&2&\\2&-1&\\3&3&\end{array}$

$D_z$ = 30

$\left\{ \begin{array}{c}x=\frac{D_x}{D}=\frac{-10}{-10} = 1 \\\\y =\frac{D_y}{D}=\frac{20}{10}=2\\\\z=\frac{D_z}{D}=\frac{30}{10}=3\\\end{array}\right.$

c) $\left[\begin{array}{ccc}2&3&1\\1&1&1\\5&2&3\end{array}\right]$

$D=\left|\begin{array}{ccc}2&3&1\\1&1&1\\5&2&3\end{array}\right| \begin{array}{ccc}2&3&\\1&1&\\5&2&\end{array}$

D = 5

Os termos independentes são 11, 6 e 18.

$D_x=\left|\begin{array}{ccc}11&3&1\\6&1&1\\18&2&3\end{array}\right| \begin{array}{ccc}11&3&\\6&1&\\18&2&\end{array}$

$D_x$ = 5

$D_y=\left|\begin{array}{ccc}2&11&1\\1&6&1\\5&18&3\end{array}\right| \begin{array}{ccc}2&11&\\1&6&\\5&18&\end{array}$

$D_y$ = 10

$D_z=\left|\begin{array}{ccc}2&3&11\\1&1&6\\5&2&18\end{array}\right| \begin{array}{ccc}2&3&\\1&1&\\5&2&\end{array}$

$D_z$ = 15

$\left\{ \begin{array}{c}x=\frac{D_x}{D}=\frac{5}{5} = 1 \\\\y =\frac{D_y}{D}=\frac{10}{5}=2\\\\z=\frac{D_z}{D}=\frac{15}{5}=3\\\end{array}\right.$