Solve Using an Inverse Matrix 6x-y+3z=-13 , x+3y-z=-4 , 3x+3y-4z=3

Problem

{[6*x−y+3*z=−13],[x+3*y−z=−4],[3*x+3*y−4*z=3])

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=[[6,−1,3],[1,3,−1],[3,3,−4]]

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

B=[[−13],[−4],[3]]

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

det(A)=6*((3)*(−4)−(−1)*(3))−(−1)*((1)*(−4)−(−1)*(3))+3*((1)*(3)−(3)*(3))

det(A)=6*(−12+3)+1*(−4+3)+3*(3−9)

det(A)=6*(−9)+1*(−1)+3*(−6)=−54−1−18=−73

3. Find the matrix of cofactors C by calculating the minor of each element.

C=[[−9,1,−6],[5,−33,−21],[−8,9,19]]

4. Determine the adjugate matrix adj(A) by taking the transpose of the cofactor matrix.

adj(A)=CT=[[−9,5,−8],[1,−33,9],[−6,−21,19]]

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

A(−1)=1/(−73)*[[−9,5,−8],[1,−33,9],[−6,−21,19]]

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

X=1/(−73)*[[(−9)*(−13)+(5)*(−4)+(−8)*(3)],[(1)*(−13)+(−33)*(−4)+(9)*(3)],[(−6)*(−13)+(−21)*(−4)+(19)*(3)]]

X=1/(−73)*[[117−20−24],[−13+132+27],[78+84+57]]

X=1/(−73)*[[73],[146],[219]]=[[−1],[−2],[−3]]

Final Answer

[[x],[y],[z]]=[[−1],[−2],[−3]]

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