Find the Inverse [[0.2,0,0.4],[0,0.9,0.05],[0,0.1,0.4]]

Problem

[[0.2,0,0.4],[0,0.9,0.05],[0,0.1,0.4]](−1)

Solution

1. Identify the matrix A and the formula for the inverse of a 3×3 matrix.

A=[[0.2,0,0.4],[0,0.9,0.05],[0,0.1,0.4]]

A(−1)=1/det(A)*adj(A)

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

det(A)=0.2*((0.9)*(0.4)−(0.05)*(0.1))−0+0

det(A)=0.2*(0.36−0.005)

det(A)=0.2*(0.355)=0.071

3. Find the matrix of cofactors (C_i*j) by calculating the minor of each element.

(C_11)=(0.9)*(0.4)−(0.05)*(0.1)=0.355

(C_12)=−(0*(0.4)−(0.05)*(0))=0

(C_13)=0*(0.1)−(0.9)*(0)=0

(C_21)=−(0*(0.4)−(0.4)*(0.1))=0.04

(C_22)=(0.2)*(0.4)−(0.4)*(0)=0.08

(C_23)=−((0.2)*(0.1)−0*(0))=−0.02

(C_31)=0*(0.05)−(0.4)*(0.9)=−0.36

(C_32)=−((0.2)*(0.05)−(0.4)*(0))=−0.01

(C_33)=(0.2)*(0.9)−0*(0)=0.18

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

adj(A)=[[0.355,0.04,−0.36],[0,0.08,−0.01],[0,−0.02,0.18]]

5. Divide each element of the adjugate matrix by the determinant 0.071

A(−1)=1/0.071*[[0.355,0.04,−0.36],[0,0.08,−0.01],[0,−0.02,0.18]]

A(−1)=[[5,40/71,−360/71],[0,80/71,−10/71],[0,−20/71,180/71]]

Final Answer

[[0.2,0,0.4],[0,0.9,0.05],[0,0.1,0.4]](−1)=[[5,40/71,−360/71],[0,80/71,−10/71],[0,−20/71,180/71]]

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