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

Problem

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

Solution

1. Write the system in matrix form A*X=B where A is the coefficient matrix, X is the variable column vector, and B is the constant column vector.

A=[[1,2,−1],[2,1,1],[1,2,1]]

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

B=[[4],[−2],[2]]

2. Calculate the determinant of A to ensure the matrix is invertible.

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

det(A)=1*(−1)−2*(1)−1*(3)

det(A)=−6

3. Find the matrix of cofactors C for matrix A

(C_11)=−1,(C_12)=−1,(C_13)=3

(C_21)=−4,(C_22)=2,(C_23)=0

(C_31)=3,(C_32)=−3,(C_33)=−3

4. Determine the adjugate matrix adj(A) by transposing the cofactor matrix.

adj(A)=[[−1,−4,3],[−1,2,−3],[3,0,−3]]

5. Compute the inverse matrix A(−1) using the formula A(−1)=1/det(A)*adj(A)

A(−1)=1/(−6)*[[−1,−4,3],[−1,2,−3],[3,0,−3]]

A(−1)=[[1/6,2/3,−1/2],[1/6,−1/3,1/2],[−1/2,0,1/2]]

6. Solve for X by multiplying A(−1) and B

X=A(−1)*B

X=[[1/6,2/3,−1/2],[1/6,−1/3,1/2],[−1/2,0,1/2]]*[[4],[−2],[2]]

x=4/6−4/3−1=2/3−4/3−3/3=−5/3

y=4/6+2/3+1=2/3+2/3+3/3=7/3

z=−2+0+1=−1

Final Answer

[[x],[y],[z]]=[[−5/3],[7/3],[−1]]

Want more problems? Check here!