Question

Solve the following system if linear equation by cramer's rule 2x-y+3z=9 x+y+z=6 x-y+z=2​

Answer 1

Answer:.............................

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Step-by-step explanation:

Answer 2

Linear equation of the cramer's rule $2x-y+3z=9$ , $x+y+z=6$ , $x-y+z=2$ is  $x=1,y=2,z=3$

Step by step solution:

Given:

$2x-y+3z=9$

$x+y+z=6$

$x-y+z=2$

To find:

$x,y,z$

Cramer Rule:

First we can find the $\;\triangle$ value

$\;\triangle=\begin{vmatrix}2&-1&3\\1&1&1\\1&-1&1\end{vmatrix}$

$\;\triangle=2(1+1)+1(1-1)+3(-1-1)$

$\;\triangle=2(2)+1(0)+3(-2)$

$\;\triangle=4-6$

$\;\triangle=-2$

then find   , $\Delta x,\Delta y,\Delta z$

$\;\triangle x=\begin{vmatrix}9&-1&3\\6&1&1\\2&-1&1\end{vmatrix}$

$\;\triangle x=9(1+1)+1(6-2)+3(-6-2)$

$\;\triangle x=9(2)+1(4)+3(-8)$

$\;\triangle x=18+4-24$

$\;\triangle x=-2$

$\;\triangle y=\begin{vmatrix}2&9&3\\1&6&1\\1&2&1\end{vmatrix}$

$\;\triangle y=2(6-2)-9(1-1)+3(2-6)$

$\;\triangle y=2(4)-9(0)+3(-4)$

$\;\triangle y=-4$

$\;\;\triangle z=\begin{vmatrix}2&-1&9\\1&1&6\\1&-1&2\end{vmatrix}$

$\;\;\triangle z=2(2+6)+1(2-6)+9(-1-1)$

$\;\;\triangle z=2(8)+1(-4)+9(-2)$

$\;\;\triangle z=-6$

finally find the values of $x. y, z$,

$\;\;x={\textstyle\frac{\triangle x}\triangle}$

$\;\;x={\textstyle\frac{-2}{-2}}$

$x=1$

$\;\;y={\textstyle\frac{\triangle y}\triangle}$

$\;\;y={\textstyle\frac{-4}{-2}}$

$y=2$

$\;\;\;z={\textstyle\frac{\triangle z}\triangle}$

$\;\;\;z={\textstyle\frac{-6}{-2}}$

$z=3$

∴$x=1 , y=2,z=3$