Question

Me pueden ayudar a resolver este problema de sel 3x3 x+3y+3z=7 2x+z=0 y+z=3​

Answer 1

Respuesta:

x = -2

y = -1

z = 4

Explicación paso a paso:

Resolviendo por determinantes:

x + 3y + 3z = 7    

2x + z = 0

y + z = 3

$x =\frac{\left[\begin{array}{ccc}7&3&3\\0&0&1\\3&1&1\end{array}\right] }{\left[\begin{array}{ccc}1&3&3\\2&0&1\\0&1&1\end{array}\right] } =\frac{[(7x0x1)+(3x1x3)+(3x0x1)]-[(3x0x3)+(1x1x7)+(1x0x3)]}{[(1x0x1)+(3x1x0)+(3x2x1)]-[(3x0x0)+(1x1x1)+(1x3x2)]} =\frac{9-7}{6-(1+6)}=\frac{2}{6-7}=\frac{2}{-1}=-2$

$y =\frac{\left[\begin{array}{ccc}1&7&3\\2&0&1\\0&3&1\end{array}\right] }{\left[\begin{array}{ccc}1&3&3\\2&0&1\\0&1&1\end{array}\right] } =\frac{[(1x0x1)+(7x1x0)+(3x2x3)]-[(3x0x0)+(1x3x1)+(1x7x2)]}{-1} =\frac{18-(3+14)}{-1}=\frac{18-17}{-1}=\frac{1}{-1}=-1$

$z =\frac{\left[\begin{array}{ccc}1&3&7\\2&0&0\\0&1&3\end{array}\right] }{\left[\begin{array}{ccc}1&3&3\\2&0&1\\0&1&1\end{array}\right] } =\frac{[(1x0x3)+(3x0x0)+(7x2x1)]-[(7x0x0)+(0x1x1)+(3x3x2)]}{-1} =\frac{14-18}{-1}=\frac{-4}{-1}=4$