Question

ii) x+y-2z = -10,
2x+y=3z = -19, 4x+6y+z = 2 by using cramers rule​

Answer 1

Given :  x+y-2z = -10,

2x+y=3z = -19,

4x+6y+z = 2

To Find : Solve using cramers rule​

Solution:

x + y - 2z = -10

2x + y + 3z  = - 19

4x  + 6y + z  = 2  

$\left|\begin{array}{ccc}1&1&-2\\2&1&3\\4&6&1\end{array}\right|= -23$

$\left|\begin{array}{ccc}-10&1&-2\\-19&1&3\\2&6&1\end{array}\right|= 427$

x = -427/23

$\left|\begin{array}{ccc}1&-10&-2\\2&-19&3\\4&2&1\end{array}\right|= -285$

=> y = 285/23

$\left|\begin{array}{ccc}1&1&-10\\2&1&-19\\4&6&2\end{array}\right|=-44$

z = 44/23

x = -427/23

y = 285/23

z = 44/23

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