Question

Resolva o sistema linar 3x3 por metodo de escalomento
(2x+3y+3z=18)
(3x+2y+5z=23 )
(5x+4y+2z=37)

Answer 1

Resposta:

x=3, y=2 e z=2

Explicação passo-a-passo:

2x+3y+3z=18

3x+2y+5z=23

5x+4y+2z=27

Vamos escrever esse sistema em forma de matriz:

$\left[\begin{array}{cccc}2&3&3&18\\3&2&5&23\\5&4&2&27\end{array}\right]$, escalonando...

$L_{1}$→$\frac{L_{1} }{2}$  $\left[\begin{array}{cccc}1&\frac{3}{2} &\frac{3}{2} &9\\3&2&5&23\\5&4&2&27\end{array}\right]$

$L_{2}$→$L_{2} -3L_{1}$ $\left[\begin{array}{cccc}1&\frac{3}{2} &\frac{3}{2} &9\\0&-\frac{5}{2} &\frac{1}{2} &-4\\5&4&2&27\end{array}\right]$

$L_{3}$→$L_{3} -5L_{1}$$\left[\begin{array}{cccc}1&\frac{3}{2} &\frac{3}{2} &9\\0&-\frac{5}{2} &\frac{1}{2} &-4\\0&-\frac{7}{2} &-\frac{11}{2} &-18\end{array}\right]$

$L_{2}$→$-\frac{2}{5} L_{2}$$\left[\begin{array}{cccc}1&\frac{3}{2} &\frac{3}{2} &9\\0&1 &-\frac{1}{5} &\frac{8}{5} \\0&-\frac{7}{2} &-\frac{11}{2} &-18\end{array}\right]$

$L_{3}$→$L_{3} +\frac{7}{2} L_{2}$ $\left[\begin{array}{cccc}1&\frac{3}{2} &\frac{3}{2} &9\\0&1 &-\frac{1}{5} &\frac{8}{5} \\0&0 &-\frac{31}{5} &-\frac{62}{5} \end{array}\right]$

$L_{1}$→$L_{1} -\frac{3}{2} L_{2}$ $\left[\begin{array}{cccc}1&0 &\frac{9}{5} &\frac{33}{5} \\0&1 &-\frac{1}{5} &\frac{8}{5} \\0&0 &-\frac{31}{5} &-\frac{62}{5} \end{array}\right]$

$L_{3}$→$-\frac{5}{31} L_{3}$ $\left[\begin{array}{cccc}1&0 &\frac{9}{5} &\frac{33}{5} \\0&1 &-\frac{1}{5} &\frac{8}{5} \\0&0 &1 &2 \end{array}\right]$

$L_{2}$→$L_{2}+\frac{1}{5} L_{3}$ $\left[\begin{array}{cccc}1&0 &\frac{9}{5} &\frac{33}{5} \\0&1 &0 &2 \\0&0 &1 &2 \end{array}\right]$

$L_{1}$→$L_{1}-\frac{9}{5} L_{3}$ $\left[\begin{array}{cccc}1&0 &0 &3 \\0&1 &0 &2 \\0&0 &1 &2 \end{array}\right]$

Portanto,

x=3

y=2

z=2