Question

Resolva o sistema abaixo utilizando a regra cramer.

(A) x + 3y - z = 0
2x + y + z = 1
3x - y + z= 3


(B) 2x + y + z= 3
-2x + 2y -z =0
3x + y + z = 1

HELP!HELP! :(
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Answer 1

$\begin{bmatrix}x+3y-z=0\\ 2x+y+z=1\\ 3x-y+z=3\end{bmatrix}\\\\\begin{bmatrix}2\left(-3y+z\right)+y+z=1\\ 3\left(-3y+z\right)-y+z=3\end{bmatrix}\\\\\begin{bmatrix}-5y+3z=1\\ -10y+4z=3\end{bmatrix}\\\\\begin{bmatrix}-10\left(-\frac{1-3z}{5}\right)+4z=3\end{bmatrix}\\\\\begin{bmatrix}-2z+2=3\end{bmatrix}\\\\y=-\frac{1-3\left(-\frac{1}{2}\right)}{5}\\\\y=-\frac{1}{2}\\\\=-3\left(-\frac{1}{2}\right)-\frac{1}{2}\\\\x=1\\\\x=1,\:z=-\frac{1}{2},\:y=-\frac{1}{2}$

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$\begin{bmatrix}2x+y+z=3\\ -2x+2y-z=0\\ 3x+y+z=1\end{bmatrix}\\\\\begin{bmatrix}-2\cdot \frac{3-y-z}{2}+2y-z=0\\ 3\cdot \frac{3-y-z}{2}+y+z=1\end{bmatrix}\\\\\begin{bmatrix}3y-3=0\\ \frac{-y-z+9}{2}=1\end{bmatrix}\\\\\begin{bmatrix}\frac{-1-z+9}{2}=1\end{bmatrix}\\\\\begin{bmatrix}\frac{-z+8}{2}=1\end{bmatrix}\\\\x=\frac{3-1-6}{2}\\\\x=-2\\\\x=-2,\:z=6,\:y=1$