Question

Resolva o sistema de três equações aplicando a regra de Cramer:​

Answer 1

Explicação passo-a-passo:

Olá!

Primeiro, monte a matriz de coeficientes:

$</p><p></p><p>\begin{bmatrix}</p><p></p><p>3 & -4 & 3 \\</p><p></p><p>2 & -1 & -1 \\</p><p></p><p>1 & -3 & -1</p><p></p><p>\end{bmatrix}$

Pela regra de Cramer:

$x=\frac{\begin{vmatrix}</p><p></p><p>-1 & -4 & 3 \\</p><p></p><p>-5 & -1 & -1 \\</p><p></p><p>-6 & -3 & -1</p><p></p><p>\end{vmatrix}}{\begin{vmatrix}</p><p></p><p>3 & -4 & 3 \\</p><p></p><p>2 & -1 & -1 \\</p><p></p><p>1 & -3 & -1</p><p></p><p>\end{vmatrix}}$

$x=-1$

$y=\frac{\begin{vmatrix}</p><p></p><p>3 & -1 & 3 \\</p><p></p><p>2 & -5 & -1 \\</p><p></p><p>1 & -6 & -1</p><p></p><p>\end{vmatrix}}{\begin{vmatrix}</p><p></p><p>3 & -4 & 3 \\</p><p></p><p>2 & -1 & -1 \\</p><p></p><p>1 & -3 & -1</p><p></p><p>\end{vmatrix}}$

$y=1$

$z=\frac{\begin{vmatrix}</p><p></p><p>3 & -4 & -1 \\</p><p></p><p>2 & -1 & -5 \\</p><p></p><p>1 & -3 & -6</p><p></p><p>\end{vmatrix}}{\begin{vmatrix}</p><p></p><p>3 & -4 & 3 \\</p><p></p><p>2 & -1 & -1 \\</p><p></p><p>1 & -3 & -1</p><p></p><p>\end{vmatrix}}$

$z=2$

Answer 2

Resposta:

$3x-4y+3z=-1\\ 2x-y-z=-5\\ x-3y-x=-6$

$D=\\ \\ \\ \left[\begin{array}{ccccc}+3&-4&+3&+3&-4\\+2&-1&-1&+2&-1\\+1&-3&-1&+1&-3\end{array}\right] \\ \\ \\ D=-11-14=-25\\ \\ D=-25$

$Dx=\\ \\ \\ \left[\begin{array}{ccccc}-1&-4&+3&-1&-4\\-5&-1&-1&-5&-1\\-6&-3&-1&-6&-3\end{array}\right] \\ \\ \\ Dx=20-(-5)=25$

$Dx=25\\ \\ \\ Dy=\\ \\ \\ \left[\begin{array}{ccccc}+3&-1&+3&+3&-1\\+2&-5&-1&+2&-5\\+1&-6&-1&+1&-6\end{array}\right] \\ \\ \\ Dy=-20-5=-25\\ \\ Dy=-25$

$Dz=\\ \\ \\ \left[\begin{array}{ccccc}+3&-4&-1&+3&-4\\+2&-1&-5&+2&-1\\+1&-3&-6&+1&-3\end{array}\right] \\ \\ \\ Dz=44-94=-50\\ \\ Dz=-50$

$x=\frac{Dx}{D} =\frac{25}{-25}=-1\\ \\ y=\frac{Dy}{D} =\frac{-25}{-25} =1\\ \\ z=\frac{Dz}{D} =\frac{-50}{-25} =2\\ \\ \\ S=\left \{ {{-1,1,2} {}} \}$