Question

necesito resolver estas ecuaciones 3x3 sea x igualación, eliminación o cramer ... por favor ayuda !!!!!

1). 2x + 6y + z =7
x + 2 y - z = -1
5 x + 7 y - 4z = 9

2). x + y + z = 0
x + 2 y + 3z = 0
3x + 5 y + 7z = 1

3). 5x - 3y - z = 1
x + 4y - 6z = -1
2x + 3y + 4z =9

Answer 1

Las soluciones de cada sistema son:

1. x=10, y = -3, z = 5
2. No tiene solucion
3. x =1, y= 1, z = 1

Explicación paso a paso:

1. $\Delta = \left|\begin{array}{ccc}2&6&1\\1&2&-1\\5&7&-4\end{array}\right| = 2(-8+7) - 6(-4+5) + 7-10 = -2-6-3=-11\\\\\\\Delta_{1} = \left|\begin{array}{ccc}7&6&1\\-1&2&-1\\9&7&-4\end{array}\right| = 7(-8+1) -6(4+9) +(-7-18)=  =-7-78-25 = -110\\\\\\\Delta_{2} = \left|\begin{array}{ccc}2&7&1\\1&-1&-1\\5&9&-4\end{array}\right| = 2(4+9) -7(-4+5) + 9+5=26-7+14=33\\\\\\\Delta_{3} = \left|\begin{array}{ccc}2&6&7\\1&2&-1\\5&7&9\end{array}\right| = 2(18+7) -6(9+5)+7(7-10) = 50 -84 -21 =-55$$x = \frac{\Delta_{1}}{\Delta} = \frac{-110}{-11} = 10\\y = \frac{\Delta_{2}}{\Delta} = \frac{33}{-11} = -3\\z = \frac{\Delta_{3}}{\Delta} = \frac{-55}{-11} = 5$
2. $\left|\begin{array}{ccc}1&1&1\\1&2&3\\3&5&7\end{array}\right| = (14-15) - (7-9) +(5-6) = -1+2-1 = 0$ Si el determinante del sistema es 0, este no tiene solucion
3. $\Delta = \left|\begin{array}{ccc}5&-3&-1\\1&4&-6\\2&3&4\end{array}\right| = 5(16+18) +3(4+12) -(3-8)=170+48+5=223\\\\\\\Delta_1 =  \left|\begin{array}{ccc}1&-3&-1\\-1&4&-6\\9&3&4\end{array}\right| = (16+18)+3(54-4)-(-3-36)=34+150+39=223\\\\\\\Delta_2 =  \left|\begin{array}{ccc}5&1&-1\\1&-1&-6\\2&9&4\end{array}\right| = 5(-4+54) -16-11 = 223\\\\\\\Delta_3 = \left|\begin{array}{ccc}5&-3&1\\1&4&-1\\2&3&9\end{array}\right| = 5( 36+3) +3*11+3-8 = 195+33-5 = 223$$x = \frac{\Delta_{1}}{\Delta}= \frac{223}{223}=1\\\\y = \frac{\Delta_{2}}{\Delta}=\frac{223}{223}=1\\\\z = \frac{\Delta_{3}}{\Delta}=\frac{223}{223}=1$