Question

Ayuda urgente
Resolver por método Cramer, con paso a paso de la solución
2x – 3y + z = 5 2x + 2y + 3z =6 -2x + 3 y = -3

Answer 1

Respuesta:

$x=\frac{3}{5}\\\\y=-\frac{3}{5}\\\\z=2$

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Explicación paso a paso:

$D =\begin{vmatrix}\begin{array}{rrr}2 & -3 & 1\\2 & 2 & 3\\-2 & 3 & 0\end{array}\end{vmatrix}=2\begin{vmatrix}\begin{array}{rr}2 & 3\\3 & 0\end{array}\end{vmatrix}-\left(-3\right)\begin{vmatrix}\begin{array}{rr}2 & 3\\-2 & 0\end{array}\end{vmatrix}+1\begin{vmatrix}\begin{array}{rr}2 & 2\\-2 & 3\end{array}\end{vmatrix}$

$&=2\left(\left(2\right)\left(0\right)-\left(3\right)\left(3\right)\right)+3\left(\left(2\right)\left(0\right)-\left(3\right)\left(-2\right)\right)+1\left(\left(2\right)\left(3\right)-\left(2\right)\left(-2\right)\right)\\&=2\left(0-9\right)+3\left(0+6\right)+1\left(6+4\right)=2\left(-9\right)+3\left(6\right)+1\left(10\right)\\&=-18+18+10\\D&=10$

$D_{x}&=\begin{vmatrix}\begin{array}{rrr}5 & -3 & 1\\6 & 2 & 3\\-3 & 3 & 0\end{array}\end{vmatrix}=5\begin{vmatrix}\begin{array}{rr}2 & 3\\3 & 0\end{array}\end{vmatrix}-\left(-3\right)\begin{vmatrix}\begin{array}{rr}6 & 3\\-3 & 0\end{array}\end{vmatrix}+1\begin{vmatrix}\begin{array}{rr}6 & 2\\-3 & 3\end{array}\end{vmatrix}\\&$

$&=5\left(\left(2\right)\left(0\right)-\left(3\right)\left(3\right)\right)+3\left(\left(6\right)\left(0\right)-\left(3\right)\left(-3\right)\right)+1\left(\left(6\right)\left(3\right)-\left(2\right)\left(-3\right)\right)\\&=5\left(0-9\right)+3\left(0+9\right)+1\left(18+6\right)=5\left(-9\right)+3\left(9\right)+1\left(24\right)\\&=-45+27+24\\D_{x}&=6$

$D_{y}&=\begin{vmatrix}\begin{array}{rrr}2 & 5 & 1\\2 & 6 & 3\\-2 & -3 & 0\end{array}\end{vmatrix}=2\begin{vmatrix}\begin{array}{rr}6 & 3\\-3 & 0\end{array}\end{vmatrix}-5\begin{vmatrix}\begin{array}{rr}2 & 3\\-2 & 0\end{array}\end{vmatrix}+1\begin{vmatrix}\begin{array}{rr}2 & 6\\-2 & -3\end{array}\end{vmatrix}$

$&=2\left(\left(6\right)\left(0\right)-\left(3\right)\left(-3\right)\right)-5\left(\left(2\right)\left(0\right)-\left(3\right)\left(-2\right)\right)+1\left(\left(2\right)\left(-3\right)-\left(6\right)\left(-2\right)\right)\\&=2\left(0+9\right)-5\left(0+6\right)+1\left(-6+12\right)=2\left(9\right)-5\left(6\right)+1\left(6\right)\\&=18-30+6\\D_{y}&=-6$

$D_{z}&=\begin{vmatrix}\begin{array}{rrr}2 & -3 & 5\\2 & 2 & 6\\-2 & 3 & -3\end{array}\end{vmatrix}=2\begin{vmatrix}\begin{array}{rr}2 & 6\\3 & -3\end{array}\end{vmatrix}-\left(-3\right)\begin{vmatrix}\begin{array}{rr}2 & 6\\-2 & -3\end{array}\end{vmatrix}+5\begin{vmatrix}\begin{array}{rr}2 & 2\\-2 & 3\end{array}\end{vmatrix}$

$&=2\left(\left(2\right)\left(-3\right)-\left(6\right)\left(3\right)\right)+3\left(\left(2\right)\left(-3\right)-\left(6\right)\left(-2\right)\right)+5\left(\left(2\right)\left(3\right)-\left(2\right)\left(-2\right)\right)\\&=2\left(-6-18\right)+3\left(-6+12\right)+5\left(6+4\right)=2\left(-24\right)+3\left(6\right)+5\left(10\right)\\&=-48+18+50\\D_{z}&=20$

$x=&\frac{D_{x}}{D}=\frac{6}{10}=\frac{3}{5}\\&\\y=&\frac{D_{y}}{D}=\frac{-6}{10}=-\frac{3}{5}\\&\\z=&\frac{D_{z}}{D}=\frac{20}{10}=2$