Question

Resolva esse sistema pela Regra de Cramer:

3x+4y-z=-1
-x+2y+z=-3
2x+5z=2

Answer 1

${\Delta}=\left|\begin{array}{ccc}3&4&-1\\-1&2&1\\2&0&5\end{array}\right| \left\begin{array}{ccc}3&4\\-1&2\\2&0\end{array}\right \\\\\\ {\Delta}=30+8-0+4-0+20~~=~~62$

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$\Delta_{x}=\left|\begin{array}{ccc}-1&4&-1\\-3&2&1\\2&0&5\end{array}\right| \left\begin{array}{ccc}-1&4\\-3&2\\2&0\end{array}\right\\\\\\ \Delta_{x}=-10+8+0+4-0+60~~=~~62$

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$\Delta_{y}= \left|\begin{array}{ccc}3&-1&-1\\-1&-3&1\\2&2&5\end{array}\right| \left\begin{array}{ccc}3&-1\\-1&-3\\2&2\end{array}\right\\\\\\ \Delta_{y}=-45-2+2-6-6-5~~=~~-62$

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$\Delta_{z}= \left|\begin{array}{ccc}3&4&-1\\-1&2&-3\\2&0&2\end{array}\right| \left\begin{array}{ccc}3&4\\-1&2\\2&0\end{array}\right\\\\\\ \Delta_{z}=12-24-0+4+0+8~~=~~0$

Pela regra de Cramer..

$x= \dfrac{\Delta_{x}}{\Delta}= \dfrac{62}{62}=1\\\\\\ y= \dfrac{\Delta_{y}}{\Delta} = \dfrac{-62}{~~62}=-1\\\\\\ z= \dfrac{\Delta_{z}}{\Delta}= \dfrac{0}{62}=0$

Portanto:

$\huge\boxed{\boxed{\text{S}=\{(1,-1,0)\}}}$