Find the Inverse [[2,4,3],[1,0,1],[3,2,2]]

Problem

[[2,4,3],[1,0,1],[3,2,2]]

Solution

1. Calculate the determinant of the matrix A using cofactor expansion along the second row.

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

det(A)=−1*(8−6)−1*(4−12)

det(A)=−2+8=6

2. Find the matrix of cofactors by calculating the determinant of the 2×2 minor for each element, applying the sign pattern.

(C_11)=+(0*(2)−1*(2))=−2

(C_12)=−(1*(2)−1*(3))=1

(C_13)=+(1*(2)−0*(3))=2

(C_21)=−(4*(2)−3*(2))=−2

(C_22)=+(2*(2)−3*(3))=−5

(C_23)=−(2*(2)−4*(3))=8

(C_31)=+(4*(1)−3*(0))=4

(C_32)=−(2*(1)−3*(1))=1

(C_33)=+(2*(0)−4*(1))=−4

3. Form the adjugate matrix by taking the transpose of the cofactor matrix.

adj(A)=[[−2,−2,4],[1,−5,1],[2,8,−4]]

4. Multiply by the reciprocal of the determinant to find the inverse matrix.

A(−1)=1/6*[[−2,−2,4],[1,−5,1],[2,8,−4]]

Final Answer

[[2,4,3],[1,0,1],[3,2,2]](−1)=[[−1/3,−1/3,2/3],[1/6,−5/6,1/6],[1/3,4/3,−2/3]]

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