Solve Using an Inverse Matrix x-3z=-2 , 2x+2y+z=4 , 3x+y-2z=5

Problem

{[x−3*z=−2],[2*x+2*y+z=4],[3*x+y−2*z=5])

Solution

1. Write the system in the 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=[[1,0,−3],[2,2,1],[3,1,−2]]

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

B=[[−2],[4],[5]]

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

det(A)=1*(2*(−2)−1*(1))−0*(2*(−2)−3*(1))+(−3)*(2*(1)−3*(2))

det(A)=1*(−5)−0+(−3)*(−4)

det(A)=−5+12=7

3. Find the adjugate matrix adj(A) by calculating the transpose of the matrix of cofactors.

(C_11)=−5,(C_12)=7,(C_13)=−4

(C_21)=−3,(C_22)=7,(C_23)=−1

(C_31)=6,(C_32)=−7,(C_33)=2

adj(A)=[[−5,−3,6],[7,7,−7],[−4,−1,2]]

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

A(−1)=1/7*[[−5,−3,6],[7,7,−7],[−4,−1,2]]

5. Solve for X by multiplying the inverse matrix by the constant vector B such that X=A(−1)*B

X=1/7*[[−5,−3,6],[7,7,−7],[−4,−1,2]]*[[−2],[4],[5]]

X=1/7*[[(−5)*(−2)+(−3)*(4)+(6)*(5)],[(7)*(−2)+(7)*(4)+(−7)*(5)],[(−4)*(−2)+(−1)*(4)+(2)*(5)]]

X=1/7*[[10−12+30],[−14+28−35],[8−4+10]]=1/7*[[28],[−21],[14]]

6. Simplify the vector to find the values of x y and z

X=[[4],[−3],[2]]

Final Answer

[[x],[y],[z]]=[[4],[−3],[2]]

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