Question

Resolva o sistema linear : x+2y+z=4

2x+y-z=1

3x-y-2z=-2

Answer 1

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

Regra de Cramer

$x=\frac{D_{x} }{D} \\\\ y=\frac{D_{y} }{D} \\ \\ z=\frac{D_{z} }{D}$

$D=\left[\begin{array}{ccccc}+1&+2&+1&+1&+2\\+2&+1&+1&+2&+1\\+3&-1&-2&+3&-1\end{array}\right] \\ \\ \\ D=-10-(-4)=-10+4=-6\\ \\ \\ D_{x} =\left[\begin{array}{ccccc}+4&+2&+1&+4&+2\\+1&+1&-1&+1&+1\\-2&-1&-2&-2&-1\end{array}\right] \\ \\ \\ D_{x} =-5-(-2)=-5+2=-3\\ \\ \\ D_{y} =\left[\begin{array}{ccccc}+1&+4&+1&+1&+4\\+2&+1&-1&+2&+1\\+3&-2&-2&+3&-2\end{array}\right] \\ \\ \\ D_{y} =-18-(-11)=-18+11=-7$

$D_{z} =\left[\begin{array}{ccccc}+1&+2&+4&+1&+2\\+2&+1&+1&+2&+1\\+3&-1&-2&+3&-1\end{array}\right] \\ \\ \\ D_{z} =-4-3=-7$

$D=-6\\ \\ D_{x} =-3\\ \\ D_{y} =-7\\\\ D_{z} =-7 \\ \\ \\ x=\frac{D_{x}}{D} =\frac{-3}{-6} =\frac{1}{2} \\ \\ y=\frac{D_{y}}{D} =\frac{-7}{-6} =\frac{7}{6} \\ \\z=\frac{D_{z}}{D} =\frac{-7}{-6} =\frac{7}{6}\\ \\ \\ S~~\left \{ {{x=\frac{1}{2};~~y=z=\frac{7}{6} } {}} \}$