Question

Solve by using Cramer’s rule

3x-y+2z=8 ; x+y+z=2 ; 2x+y-z=-1​

Answer 1

The values of x , y and z are 1 , -1 , 2 respectively.

Step-by-step explanation:

Given equations are :

3x - y + 2z = 8

x + y + z = 2

2x + y - z = -1

=> We know that Cramer's rule gives the following mathematical expressions :

$x=\frac{D_{x} }{D} \\\\y=\frac{D_{y} }{D} \\\\z=\frac{D_{z} }{D}$

=> Writing the above equation in AX=B form,

$\left[\begin{array}{ccc}3&-1&2\\1&1&1\\2&1&-1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}8\\2\\-1\end{array}\right]$

=> Finding the values of $D , D_{x}, D_{y} ,D_{z}$

$D = \left[\begin{array}{ccc}3&-1&2\\1&1&1\\2&1&-1\end{array}\right]$

D = 3(-1 -1) - (-1) (-1-2) + 2 (1-2)

D = 3(-2) + 1(-3) + 2(-1)

D = -6 -3 -2

D = - 11

$D_{x} = \left[\begin{array}{ccc}8&-1&2\\2&1&1\\-1&1&-1\end{array}\right]\\\\D_{x}= 8(-1-1) -(-1)(-2+1)+2(2+1)\\D_{x} = 8(-2)+1(-1)+2(3)\\D_{x} = -16 -1+6\\D_{x} = -11$

$D_{y} = \left[\begin{array}{ccc}3&8&2\\1&2&1\\2&-1&-1\end{array}\right]\\\\D_{y} = 3(-2+1)-8(-1-2)+2(-1-4)\\D_{y} = 3(-1)-8(-3)+2(-5)\\D_{y} = -3+24-10\\D_{y} = 11$

$D_{z} = \left[\begin{array}{ccc}3&-1&8\\1&1&2\\2&1&-1\end{array}\right]\\\\D_{z} = 3(-1-2)-(-1)(-1-4)+8(1-2)\\D_{z} = 3(-3)+1(-5)+8(-1)\\D_{z} = -9-5-8\\D_{z} = -22$

=> Using Cramer's rule to find the values of x,y and z

$- > x = \frac{D_{x} }{D}\\\\ x = \frac{-11}{-11}\\\\ x = 1\\\\\\y = \frac{D_{y} }{D}\\\\y = \frac{11}{-11}\\\\ y = -1\\\\\\z = \frac{D_{z} }{D}\\\\z= \frac{-22}{-11}\\\\ z =2$

Hence the values of x , y and z are 1 , -1 , 2 respectively.