Solve Using an Inverse Matrix x+y+xy=19 , y+z+yz=11 , z+x+zx=14

Problem

{[x+y+x*y=19],[y+z+y*z=11],[z+x+z*x=14])

Solution

1. Transform the equations by adding 1 to both sides of each equation to facilitate factoring.

x+y+x*y+1=19+1

y+z+y*z+1=11+1

z+x+z*x+1=14+1

2. Factor each equation into a product of binomials.

(x+1)*(y+1)=20

(y+1)*(z+1)=12

(z+1)*(x+1)=15

3. Substitute variables to linearize the system by taking the natural logarithm of both sides. Let a=ln(x+1) b=ln(y+1) and c=ln(z+1)

a+b=ln(20)

b+c=ln(12)

c+a=ln(15)

4. Represent the linear system in matrix form A*X=B

[[1,1,0],[0,1,1],[1,0,1]]*[[a],[b],[c]]=[[ln(20)],[ln(12)],[ln(15)]]

5. Calculate the inverse of matrix A

A(−1)=[[0.5,−0.5,0.5],[0.5,0.5,−0.5],[−0.5,0.5,0.5]]

6. Solve for the vector X using X=A(−1)*B

a=0.5*(ln(20)−ln(12)+ln(15))=ln(√(,(20⋅15)/12))=ln(5)

b=0.5*(ln(20)+ln(12)−ln(15))=ln(√(,(20⋅12)/15))=ln(4)

c=0.5*(−ln(20)+ln(12)+ln(15))=ln(√(,(12⋅15)/20))=ln(3)

7. Back-substitute to find the values of x y and z

x+1=ea=5⇒x=4

y+1=eb=4⇒y=3

z+1=ec=3⇒z=2

Final Answer

(x,y,z)=(4,3,2)

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