Solve Using an Inverse Matrix 2x-y=4 , 3x+y-z=10 , y+z=3

Problem

{[2*x−y=4],[3*x+y−z=10],[y+z=3])

Solution

1. Write the system as a matrix equation of the form A*X=B

A=[[2,−1,0],[3,1,−1],[0,1,1]]

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

B=[[4],[10],[3]]

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

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

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

det(A)=7

3. Find the matrix of cofactors for A

(C_11)=2,(C_12)=−3,(C_13)=3

(C_21)=1,(C_22)=2,(C_23)=−2

(C_31)=1,(C_32)=2,(C_33)=5

4. Determine the adjugate matrix by transposing the cofactor matrix.

adj(A)=[[2,1,1],[−3,2,2],[3,−2,5]]

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

A(−1)=1/7*[[2,1,1],[−3,2,2],[3,−2,5]]

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

X=A(−1)*B

X=1/7*[[2*(4)+1*(10)+1*(3)],[−3*(4)+2*(10)+2*(3)],[3*(4)−2*(10)+5*(3)]]

X=1/7*[[21],[14],[7]]

X=[[3],[2],[1]]

Final Answer

[[x],[y],[z]]=[[3],[2],[1]]

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