Solve Using an Inverse Matrix x-y-z=0 , 0-2y-z=-1 , -4x-5y=-2

Problem

{[x−y−z=0],[−2*y−z=−1],[−4*x−5*y=−2])

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,−1,−1],[0,−2,−1],[−4,−5,0]]

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

B=[[0],[−1],[−2]]

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

det(A)=1*((−2)*(0)−(−1)*(−5))−(−1)*((0)*(0)−(−1)*(−4))+(−1)*((0)*(−5)−(−2)*(−4))

det(A)=1*(−5)+1*(−4)−1*(−8)

det(A)=−5−4+8=−1

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

(C_11)=−5,(C_12)=4,(C_13)=−8

(C_21)=5,(C_22)=−4,(C_23)=9

(C_31)=−1,(C_32)=1,(C_33)=−2

adj(A)=[[−5,5,−1],[4,−4,1],[−8,9,−2]]

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

A(−1)=1/(−1)*[[−5,5,−1],[4,−4,1],[−8,9,−2]]

A(−1)=[[5,−5,1],[−4,4,−1],[8,−9,2]]

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

X=A(−1)*B=[[5,−5,1],[−4,4,−1],[8,−9,2]]*[[0],[−1],[−2]]

x=(5)*(0)+(−5)*(−1)+(1)*(−2)=3

y=(−4)*(0)+(4)*(−1)+(−1)*(−2)=−2

z=(8)*(0)+(−9)*(−1)+(2)*(−2)=5

Final Answer

[[x],[y],[z]]=[[3],[−2],[5]]

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