Find the Inverse

Problem

[[1,−3,0,−2],[3,−12,−2,−6],[−2,10,2,5],[−1,6,1,3]](−1)

Solution

1. Set up the augmented matrix [A|I] by placing the identity matrix (I_4) to the right of the given matrix A

[[1,−3,0,−2,|,1,0,0,0],[3,−12,−2,−6,|,0,1,0,0],[−2,10,2,5,|,0,0,1,0],[−1,6,1,3,|,0,0,0,1]]

2. Eliminate the first column entries below the pivot in the first row using row operations (R_2)→(R_2)−3*(R_1) (R_3)→(R_3)+2*(R_1) and (R_4)→(R_4)+(R_1)

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

3. Normalize the second row by multiplying (R_2) by −1/3

[[1,−3,0,−2,|,1,0,0,0],[0,1,2/3,0,|,1,−1/3,0,0],[0,4,2,1,|,2,0,1,0],[0,3,1,1,|,1,0,0,1]]

4. Eliminate the second column entries around the pivot in the second row using (R_1)→(R_1)+3*(R_2) (R_3)→(R_3)−4*(R_2) and (R_4)→(R_4)−3*(R_2)

[[1,0,2,−2,|,4,−1,0,0],[0,1,2/3,0,|,1,−1/3,0,0],[0,0,−2/3,1,|,−2,4/3,1,0],[0,0,−1,1,|,−2,1,0,1]]

5. Normalize the third row by multiplying (R_3) by −3/2

[[1,0,2,−2,|,4,−1,0,0],[0,1,2/3,0,|,1,−1/3,0,0],[0,0,1,−3/2,|,3,−2,−3/2,0],[0,0,−1,1,|,−2,1,0,1]]

6. Eliminate the third column entries around the pivot in the third row using (R_1)→(R_1)−2*(R_3) (R_2)→(R_2)−2/3*(R_3) and (R_4)→(R_4)+(R_3)

[[1,0,0,1,|,−2,3,3,0],[0,1,0,1,|,−1,1,1,0],[0,0,1,−3/2,|,3,−2,−3/2,0],[0,0,0,−1/2,|,1,−1,−3/2,1]]

7. Normalize the fourth row by multiplying (R_4) by −2

[[1,0,0,1,|,−2,3,3,0],[0,1,0,1,|,−1,1,1,0],[0,0,1,−3/2,|,3,−2,−3/2,0],[0,0,0,1,|,−2,2,3,−2]]

8. Eliminate the fourth column entries above the pivot in the fourth row using (R_1)→(R_1)−(R_4) (R_2)→(R_2)−(R_4) and (R_3)→(R_3)+3/2*(R_4)

[[1,0,0,0,|,0,1,0,2],[0,1,0,0,|,1,−1,−2,2],[0,0,1,0,|,0,1,3,−3],[0,0,0,1,|,−2,2,3,−2]]

Final Answer

[[1,−3,0,−2],[3,−12,−2,−6],[−2,10,2,5],[−1,6,1,3]](−1)=[[0,1,0,2],[1,−1,−2,2],[0,1,3,−3],[−2,2,3,−2]]

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