Solve Using an Inverse Matrix x+y+z=12 , 2x-3y+2z=4 , x+z=2y

Problem

{[x+y+z=12],[2*x−3*y+2*z=4],[x+z=2*y])

Solution

1. Rearrange the third equation into standard form a*x+b*y+c*z=d by subtracting 2*y from both sides to get x−2*y+z=0

2. Write the system as a matrix equation A*X=B where A is the coefficient matrix, X is the column vector of variables, and B is the column vector of constants.

A=[[1,1,1],[2,−3,2],[1,−2,1]]

X=[[x],[y],[z]]

B=[[12],[4],[0]]

3. Calculate the determinant of matrix A denoted as det(A) using the first row.

det(A)=1*((−3)*(1)−(2)*(−2))−1*((2)*(1)−(2)*(1))+1*((2)*(−2)−(−3)*(1))

det(A)=1*(1)−1*(0)+1*(−1)

det(A)=0

4. Conclude the status of the system. Since the determinant of the coefficient matrix A is 0 the matrix is singular and does not have an inverse. This means the system either has no solution or infinitely many solutions.

5. Analyze the equations further. Notice that the first equation x+y+z=12 and the third equation x−2*y+z=0 can be subtracted to eliminate x and z

(x+y+z)−(x−2*y+z)=12−0

3*y=12

y=4

6. Substitute y=4 into the first and second equations.

x+4+z=12⇒x+z=8

2*x−3*(4)+2*z=4⇒2*x+2*z=16⇒x+z=8

7. Determine the final solution set. Since both remaining equations simplify to x+z=8 there are infinitely many solutions. Let z=t

x=8−t

y=4

z=t

Final Answer

The system has infinitely many solutions: *(x,y,z)=(8−t,4,t)

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