Question

RECOMPENSA: 40 pontos

Gincana da tarde.

Utilizando a Regra de Cramer, determine o valor da incógnita y no seguinte sistema de equações lineares: 

Boa sorte ksksk.​

Answer 1

Olá, boa tarde!

Resposta:

$\Large{\color{blue}{\boxed{\boxed{\boxed{\mathsf{\color{red}{S = \left\{x=3; y=2; z=2 \right\}}}}}}}}$

Explicação passo-a-passo:

Usando a Regra de Cramer, devemos representar o seguinte sistema apresentado na imagem acima em forma de Matriz.

$\mathsf{\left[\begin{array}{ccc}2 & 3 & 3 \\ 3 & 2 & 5 \\ 5 & 4 & 2 \end{array}\right] \ \left[\begin{array}{ccc}x \\ y \\ z \end{array}\right] \ = \left[\begin{array}{ccc}18 \\ 23 \\ 27 \end{array}\right]}$

✓ Determinante de cada Matriz

$\mathsf{\left[\begin{array}{ccc}2 & 3 & 3 \\ 3 & 2 & 5 \\ 5 & 4 & 2 \end{array}\right]}$

$\mathsf{D = 2 \left[\begin{array}{cc}2 & 5 \\ 4 & 2 \end{array}\right] -3 \left[\begin{array}{cc}3 & 3 \\ 4 & 2 \end{array}\right] +5 \left[\begin{array}{cc} 3 & 3 \\ 2 & 5 \end{array}\right]}$

Fórmula das matrizes

ad - cb

° 2 • (2.2 - 5.4) = - 32

° - 3 • (2.3 - 4.3) = 18

° 5 • (3.5 - 2.3) = 45

$\mathsf{D = -32+18+45}$

$\boxed{\mathsf{D=31}}$

=> Determinante de x

$\mathsf{\left[\begin{array}{ccc}18 & 3 & 3 \\ 23 & 2 & 5 \\ 27 & 4 & 2 \end{array}\right]}$

$\mathsf{D_x = 18 \left[\begin{array}{cc}2 & 5 \\ 4 & 2 \end{array}\right] -23 \left[\begin{array}{cc} 3 & 3 \\ 4 & 2 \end{array}\right] +27 \left[\begin{array}{cc} 3 & 3 \\ 2 & 5 \end{array}\right]}$

Fórmula de matrizes

ad - cb

° 18 • (2.2 - 4.5) = - 288

° - 23 • (2.3 - 4.3) = 138

° 27 • (5.3 - 2.3) = 243

$\mathsf{D_x = -288+138+243}$

$\boxed{\mathsf{D_x=93}}$

=> Determinante de y

$\mathsf{\left[\begin{array}{ccc}2 & 18 & 3 \\ 3 & 23 & 5 \\ 5 & 27 & 2\end{array}\right]}$

$\mathsf{D_y = 2 \left[\begin{array}{cc}23 & 5 \\ 27 & 2 \end{array}\right] -3 \left[\begin{array}{cc} 18 & 3 \\ 27 & 2 \end{array}\right] +5 \left[\begin{array}{cc}18 & 3 \\ 23 & 5 \end{array}\right]}$

Fórmula das matrizes

ad - bc

° 2 • (2.23 - 27.5) = - 178

° - 3 • (2.18 - 27.3) = 135

° 5 • (5.18 - 23.3) = 105

$\mathsf{D_y=-178+135+105}$

$\boxed{\mathsf{D_y=62}}$

=> Determinante de z

$\mathsf{\left[\begin{array}{ccc}2 & 3 & 18 \\ 3 & 2 & 23 \\ 5 & 4 & 27 \end{array}\right]}$

$\mathsf{D_z = 2 \left[\begin{array}{cc}2 & 23 \\ 4 & 27 \end{array}\right] -3 \left[\begin{array}{cc}3 & 18 \\ 4 & 27 \end{array}\right] +5 \left[\begin{array}{cc}3 & 18 \\ 2 & 23 \end{array}\right]}$

Fórmula das matrizes

ad - bc

° 2 • (27.2 - 4.23) = - 76

° - 3 • (27.3 - 4.18) = - 27

° 5 • (23.3 - 2.18) = 165

$\mathsf{D_z=-76-27+165}$

$\boxed{\mathsf{D_z=62}}$

E finalmente podemos encontrar os valores de cada variável, usando a famosa Regra de Cramer:

✓ Valor de x = 3

$\mathsf{x = \dfrac{D_x}{D}}$

$\mathsf{x= \dfrac{93}{31}}$

$\boxed{\mathsf{x=3}}$

✓ Valor de y = 2

$\mathsf{y= \dfrac{D_y}{D}}$

$\mathsf{y= \dfrac{62}{31}}$

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

✓ Valor de z = 2

$\mathsf{z= \dfrac{D_z}{D}}$

$\mathsf{z= \dfrac{62}{31}}$

$\boxed{\mathsf{z=2}}$

Conjunto Solução:

$\boxed{\mathsf{S = \left\{x=3; y=2; z=2 \right\}}}$

Answer 2

Resposta:

y= 2

Explicação passo-a-passo:

Usando o Método de Cramer :

( 2 3 3)(2 3)

(3 2 5)(3 2)

(5 4 2)(5 4)

D=8+75+36-(30+40+18)

D=83+36-(70+18)

D=119-(88)

D=119-88

D=31

(2 18 3)(2 18)

(3 23 5)(3 23)

(5 27 2)(5 27)

Dy=92+450+243-(345+270+108)

Dy=785-(723)

Dy=785-723

Dy=62

Y=Dy/D=62/31=2