Question

inverse by using elementary column transfer
[ 1 0 1
0 2 3
1 2 1 ]​

Answer 1

Answer:

Let A = `[(0,1,2),(1,2,3),(3,1,1)]`

|A| = 0[2-3]-1[1-9] + 2[1-6]

`= 8 + 2(-5) = 8 - 10 = -2 != 0 `

`=> A^(-1)` exists

`:. A = IA`

`[(0,1,2),(1,2,3),(3,1,1)] = [(1,0,0),(0,1,0),(0,0,1)]A`

`R_1 ↔ R_2`

`[(0,1,2),(1,2,3),(3,1,1)] = [(1,0,0),(0,1,0),(0,0,1)]A`

`R_3"/"R_3 - 3R_3`

`[(1,2,3),(0,1,2),(0,-5,-8)] = [(0,1,0),(1,0,0),(0,-3,1)]A`

`R_3"/"R_3 + 5R_2`

`[(1,2,3),(0,1,2),(0,0,2)] = [(0,1,0),(1,0,0),(5, -3, 1)]A`

`R_2 "/"R_2 - R_3`

`[(1,2,3),(0,1,0),(0,0,2)] = [(0,1,0),(-4,3,-1),(5,-3,1)]A`

`R_1"/"R_1 -2R_2` and `R_3 xx 1/2`

`[(1,0,3),(0,1,0),(0,0,1)] = [(8,-5,2),(-4,3,-1),(5/3, (-3)/2, 1/2)]A`

`R_1"/"R_1 - 3R_3`

`[(1,0,0),(0,1,0),(0,0,1)] = [(1/2, -1/2 , 1/2),(-4, 3,-1),(5/2, (-3)/2 , 1/2)]A`

`:. A^(-1) = 1/2[(1,-1,1),(-8, 6, -2), (5,-3,1)]`

Concept: Matrices - Proof of the Uniqueness of Inverse