Solve Using an Inverse Matrix -2x+2y+3z=1 , x-y=3 , y+4z=-2

Problem

{[−2*x+2*y+3*z=1],[x−y=3],[y+4*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=[[−2,2,3],[1,−1,0],[0,1,4]]

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

B=[[1],[3],[−2]]

2. Calculate the determinant of matrix A using expansion along the first column.

det(A)=−2*((−1)*(4)−(0)*(1))−1*((2)*(4)−(3)*(1))+0

det(A)=−2*(−4)−1*(5)

det(A)=8−5=3

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

(C_11)=−4,(C_12)=−4,(C_13)=1

(C_21)=−5,(C_22)=−8,(C_23)=2

(C_31)=3,(C_32)=3,(C_33)=0

C=[[−4,−4,1],[−5,−8,2],[3,3,0]]

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

adj(A)=CT=[[−4,−5,3],[−4,−8,3],[1,2,0]]

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

A(−1)=1/3*[[−4,−5,3],[−4,−8,3],[1,2,0]]

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

X=1/3*[[−4,−5,3],[−4,−8,3],[1,2,0]]*[[1],[3],[−2]]

X=1/3*[[(−4)*(1)+(−5)*(3)+(3)*(−2)],[(−4)*(1)+(−8)*(3)+(3)*(−2)],[(1)*(1)+(2)*(3)+(0)*(−2)]]

X=1/3*[[−4−15−6],[−4−24−6],[1+6+0]]=1/3*[[−25],[−34],[7]]

Final Answer

[[x],[y],[z]]=[[−25/3],[−34/3],[7/3]]

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